Find angle UFO in the picture attached
Consider a regular 36-gon $A_1A_2\ldots A_{36}$ inscribed in a circle of radius $R$. Inscribed angle over any side is $5^\circ$. We can see our configuration as it is shown on the picture.
It suffices to prove that $UF$ is parallel to the diagonal $A_{13}A_{34}=EA_{34}$; then we have $\angle NFU=\angle NEA_{34}=25^\circ$, so $\angle UFO= \angle NFO-\angle NFU=100^\circ-25^\circ=75^\circ$.
To prove $UF\parallel A_{34}E$, it is enough to prove $\frac{NU}{UT}=\frac{NF}{FE}$ ($T$ is as on the picture). For we can use the following two formulae:
The length of a chord of a circle with inscribed angle $\alpha$ is $2R\sin\alpha$.
If $E$ is on a side $BC$ of a $\triangle ABC$, then $\frac{BE}{EC}= \frac{AB\sin\angle BAE}{AC\sin\angle CAE}$.
Now, from $\triangle OEN$ we have: $$\frac{NF}{FE}= \frac{ON\sin\angle NOF}{OE\sin\angle EOF}=\frac{2R\sin 40^\circ\sin 20^\circ}{2R\sin 60^\circ\sin 60^\circ}.$$ From $\triangle NET$ we have: $$\frac{NU}{UT}= \frac{EN\sin\angle NEU}{ET\sin\angle TEU}= \frac{2R\sin 80^\circ\sin 5^\circ}{ET\sin 20^\circ}.$$ By the law of sines on $\triangle NET$, $\frac{ET}{NE}=\frac{\sin 60^\circ}{\sin 95^\circ}$, so $ET= NE\ \frac{\sin 60^\circ}{\sin 95^\circ}= 2R\sin 80^\circ\frac{\sin 60^\circ}{\sin 95^\circ}$ and thus $$\frac{NU}{UT}= \frac{2R\sin 80^\circ\sin 5^\circ}{2R\sin 80^\circ\frac{\sin 60^\circ}{\sin 95^\circ}\sin 20^\circ}= \frac{2R\sin 95^\circ\sin 5^\circ}{2R\sin 60^\circ\sin 20^\circ}.$$
So, for $\frac{NU}{UT}=\frac{NF}{FE}$ it is enough to check: $\sin 40^\circ\sin 20^\circ\sin 20^\circ= \sin 95^\circ\sin 5^\circ\sin 60^\circ$.
We have: $$\sin 95^\circ\sin 5^\circ\sin 60^\circ=\frac{1}{2}(\cos 90^\circ-\cos 100^\circ)\sin 60^\circ= \frac{1}{2}\cos 80^\circ\sin 60^\circ= \frac{1}{4}(\sin 140^\circ-\sin 20^\circ)= \frac{1}{4}(\sin 40^\circ-\sin 20^\circ),$$ and: $$\sin 40^\circ\sin 20^\circ\sin 20^\circ=\frac{1}{2}(\cos 20^\circ-\cos 60^\circ)\sin 20^\circ= \frac{1}{2}(\cos 20^\circ\sin 20^\circ-\frac{1}{2}\sin 20^\circ)= \frac{1}{2}(\frac{1}{2}\sin 40^\circ-\frac{1}{2}\sin 20^\circ)= \frac{1}{4}(\sin 40^\circ-\sin 20^\circ).$$
Without loss of generality, let $OE=1$. By the Law of Sines on the triangle $ONE$, $ON=\dfrac{\sin(40^\circ)}{\sin(60^\circ)}$. Thus, using the Law of Sines on the triangle $ONF$, we get $$NF=ON\,\left(\frac{\sin(20^\circ)}{\sin(100^\circ)}\right)=ON\,\left(\frac{\sin(20^\circ)}{\sin(80^\circ)}\right)=\frac{\sin(20^\circ)\,\sin(40^\circ)}{\sin(60^\circ)\,\sin(80^\circ)}\,.$$ Furthermore, the Law of Sines on the triangle $OUE$ gives $$OU=\frac{\sin(35^\circ)}{\sin(65^\circ)}\,.$$ We also have $NE=\dfrac{\sin(80^\circ)}{\sin(60^\circ)}$ (applying the Law of Sines on the triangle $ONE$), which gives $$NU=NE\,\left(\frac{\sin(5^\circ)}{\sin(115^\circ)}\right)=NE\,\left(\frac{\sin(5^\circ)}{\sin(65^\circ)}\right)=\frac{\sin(80^\circ)\,\sin(5^\circ)}{\sin(60^\circ)\,\sin(65^\circ)}\,,$$ using the Law of Sines on the triangle $UNE$.
Thus, $$\frac{NU}{NF}=\frac{\sin^2(80^\circ)\,\sin(5^\circ)}{\sin(20^\circ)\,\sin(40^\circ)\,\sin(65^\circ)}\,.\tag{*}$$ Note that $$\sin(65^\circ)\,\sin(25^\circ)=\frac{1}{2}\,\big(\cos(40^\circ)-\cos(90^\circ)\big)=\frac{\cos(40^\circ)}{2}\,,$$ where we use the identity $\sin(x)\,\sin(y)=\dfrac{1}{2}\,\big(\cos(x-y)-\cos(x+y)\big)$. Thus, (*) becomes $$\frac{NU}{NF}=\frac{2\,\sin^2(80^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)\,\sin(40^\circ)\,\cos(40^\circ)}\,.$$ From the identity $\sin(2x)=2\,\sin(x)\,\cos(x)$, we get $$\frac{NU}{NF}=\frac{4\,\sin(80^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)}=\frac{4\,\cos(10^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)}\,.$$ That is, $$\frac{NU}{NF}=\frac{4\,\cos(10^\circ)\,\cos(5^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)\,\cos(5^\circ)}\,.$$ As $\sin(4x)=2\,\sin(2x)\,\cos(2x)=4\,\sin(x)\,\cos(x)\,\cos(2x)$, we get $$\frac{NU}{NF}=\frac{\sin(25^\circ)}{\cos(5^\circ)}=\frac{\sin(25^\circ)}{\sin(95^\circ)}\,.$$ Ergo, if $\theta:=\angle UFN$, then we have from the Law of Sines on the triangle $UNF$ that $$\frac{\sin(\theta)}{\sin(120^\circ-\theta)}=\frac{NU}{NF}=\frac{\sin(25^\circ)}{\sin(120^\circ-25^\circ)}\,.$$ It follows immediately from the identity $\sin(x)\,\sin(y)=\dfrac{1}{2}\,\big(\cos(x-y)-\cos(x+y)\big)$ that $$\cos(120^\circ+25^\circ-\theta)=\cos(120^\circ-25^\circ+\theta)\,.$$ That is, $$25^\circ-\theta=n\cdot 180^\circ$$ for some integer $n$. As $0^\circ<\theta<100^\circ$ (because $\angle OFN=100^\circ$), we have $n=0$, whence $\theta=25^\circ$. That is, $$\angle UFO=180^\circ-25^\circ-80^\circ=75^\circ\,.$$
The angles alone do not suffice. We can find out all except $\angle UFO$, $\angle NFU$, $\angle EUF$, or $\angle FUN$; and although we know that $\angle UFO + \angle NFU = 100°$ and $\angle EUF + \angle FUN = 115°$, they are not sufficient. (Even the quadrilateral $UFEO$ does not help, since it just repeats what you find from $\triangle UFO$.)
I think the trick might be to add point $G$ between $O$ and $U$ such that $\overline{GE} \parallel \overline{UF}$. If we call the point where $\overline{OF}$ and $\overline{UE}$ intersect $X$, and the point where $\overline{OF}$ and $\overline{GE}$ intersect $Y$, triangles $\triangle UFX$ and $\triangle OGY$ are similar:
$$\begin{aligned}
\angle FUX &= \angle GOY = \angle NOF = 20° \\
\angle FXU &= \angle OXE = 180° - 60° - 35° = 85° \\
\angle UFO &= 180° - \angle FUX - \angle FXU = 75° \\
\end{aligned}$$
Without any loss of generality, we can choose a Cartesian coordinate system where $O$ is at origin, $F$ is on the positive $y$ axis at $(0, 1)$, and $E$ is at $(E_x , E_y)$ with $E_x \gt 0$.
Because the angles in $\triangle NEO$ sum to $180°$, $\angle ONE = 60°$, and $\angle EON = 60° + 20° = 80°$, we know that $\angle NEO = 180° - 60° - 80° = 40°$. Because $\angle UEO = 35°$, $\angle NEU = 40° - 35° = 5°$.
Similarly, using $\triangle FOE$, we know that $\angle OFE = 80°$.
The slope of $OE$ is $30°$ (because $\angle FOE = 60°$), and the slope of $FE = -10°$ (because $\angle OFE = 80°$). This allows us to find the location of $E$: $$\begin{aligned} y_{OE}(x) &= x \tan(30°) \\ y_{FE}(x) &= 1 - x \tan(10°) \\ y_{OE}(x) &= y_{FE}(x) \end{aligned} \quad \iff \quad x = \frac{1}{\tan\left(\frac{\pi}{18}\right) + \sqrt{\frac{1}{3}}}$$ Substituting into $y_{OE}(x)$ or $y_{FE}(x)$ we get the $y$ coordinate. Thus, $$\left\lbrace\begin{aligned} E_x &= \frac{1}{\tan\left(\frac{\pi}{18}\right) + \sqrt{\frac{1}{3}}} \approx 1.326827896 \\ E_y &= \frac{1}{\sqrt{3}\tan\left(\frac{\pi}{18}\right) + 1} \approx 0.766044443 \\ \end{aligned}\right.$$
The slope of $OU$ is $70°$ (because $\angle NOF = 20°$), and the slope of $EU$ is $5°$ (because $\angle UEO = 35°$, $5°$ more than the slope of $OE$). Using these, we can solve the location of $U = (U_x , U_y)$ (noting that $U_x \lt 0$): $$\begin{aligned} y_{EU}(x) &= E_y + (E_x - x)\tan(5°) \\ y_{OU}(x) &= -x \tan(70°) \\ y_{EU}(x) &= y_{OU}(x) \end{aligned}$$ i.e. $$\left\lbrace\begin{aligned} U_x &= \frac{E_y + E_x \tan\left(\frac{\pi}{36}\right)}{\tan\left(\frac{\pi}{36}\right) - \tan\left(\frac{7\pi}{18}\right)} \approx -0.33162803 \\ U_y &= -\tan\left(\frac{7\pi}{18}\right) U_x \approx 0.91114054 \\ \end{aligned}\right.$$ The $\angle UFO$ fulfills $$\tan\left(\angle UFO\right) = \frac{-U_x}{1 - U_y} \quad \iff \quad \angle UFO = \arctan\left(\frac{U_x}{U_y - 1}\right)$$ which plugging in to a symbolic calculator (I used Maple) does not simplify to anything simple, but numerically yields $$\angle UFO = 75.000000°$$
Using $\triangle FON$, we know $\angle NFO = 100°$. Since $\angle UFO = 75°$, $\angle NFU = 25°$. Because the slope of the line $EU$ is $5°$ and $OF$ is vertical, we know $\angle UXF = \angle OXE = 85°$, and $\angle OXU = \angle FXE = 95°$. Using $\triangle NEU$, we know $\angle EUN = 115°$. Using $\triangle XUF$, we know $\angle XUF = 180° - 75° - 85° = 20°$.
In other words, the angles in the $\triangle XUF$ are $85°$, $20°$, and $75°$, respectively, so there are no similar triangles in the figure we could have used.
Furthermore, while the angle between $x$ axis and line $OE$ is $30°$, the angle between $x$ axis and line $UF$ is $5° + 20° = 25°$, so those two lines are not parallel, either.
I suspect this was one of those puzzles that try to trick you, XKCD style, while the correct answer requires just some work and care for detail; no shortcuts.
Here's how saulspatz' answer implements the calculation.
Without loss of generality, choose a Cartesian coordinate system where $O$ is at origin, and $E$ is at $(1,0)$. All the other points have a positive $y$ coordinate.
Two lines intersecting above the $x$ axis, one through $O$ with angle $\phi_1$ to the $x$ axis, and the other through $E$ with angle $\phi_2$ to the $x$ axis, intersect at $(x, y)$: $$\left\lbrace\begin{aligned} y_1(x) &= x \tan(\phi_1) \\ y_2(x) &= (1 - x) \tan(\phi_2) \\ y_1(x) &= y_2(x) \\ \end{aligned}\right. \quad \implies \quad \left\lbrace\begin{aligned} x &= \frac{\tan(\phi_2)}{\tan(\phi_1) + \tan(\phi_2)} \\ y &= \frac{\tan(\phi_1)\tan(\phi_2)}{\tan(\phi_1) + \tan(\phi_2)} \\ \end{aligned}\right. \tag{1}\label{NA1}$$
We can use $\eqref{NA1}$ to find both $U$ and $F$. For $U$, $\phi_1 = 80° = \frac{4}{9}\pi$ and $\phi_2 = 35° = \frac{7}{36}\pi$. Thus, $$\left\lbrace\begin{aligned} U_x &= \frac{\tan\left(\frac{7\pi}{36}\right)}{\tan\left(\frac{4\pi}{9}\right) + \tan\left(\frac{7\pi}{36}\right)} \approx 0.109897 \\ U_y &= \frac{\tan\left(\frac{4\pi}{9}\right) \tan\left(\frac{7\pi}{36}\right)}{\tan\left(\frac{4\pi}{9}\right) + \tan\left(\frac{7\pi}{36}\right)} \approx 0.623257 \\ \end{aligned}\right.$$ For $F$, $\phi_1 = 60° = \frac{1}{3}\pi$ and $\phi_2 = 40° = \frac{2}{9}\pi$ (because $\angle NEO = 180°-60°-20°-60° = 40°$): $$\left\lbrace\begin{aligned} F_x &= \frac{\tan\left(\frac{2\pi}{9}\right)}{\tan\left(\frac{\pi}{3}\right) + \tan\left(\frac{2\pi}{9}\right)} \\ F_y &= \frac{\tan\left(\frac{\pi}{3}\right) \tan\left(\frac{2\pi}{9}\right)}{\tan\left(\frac{\pi}{3}\right) + \tan\left(\frac{2\pi}{9}\right)} \\ \end{aligned}\right.$$
Next, we can use basic vector algebra, $$\cos\left(\angle XYZ\right) = \frac{\overline{XY} \cdot \overline{ZY}}{\left\lVert\overline{XY}\right\rVert \left\lVert\overline{ZY}\right\rVert} = \frac{\overline{XY} \cdot \overline{ZY}}{\sqrt{\left(\overline{XY}\cdot\overline{XY}\right)\left(\overline{ZY}\cdot\overline{ZY}\right)}}$$ To solve the angle $\angle UFO$, we need vectors $\overline{UF}$ and $\overline{OF}$: $$\left\lbrace\begin{aligned} \overline{UF} &= \left [ \begin{matrix} F_x - U_x \\ F_y - U_y \end{matrix} \right ] \\ \overline{OF} &= \left [ \begin{matrix} F_x \\ F_y \end{matrix} \right ] \\ \end{aligned}\right.$$ This means the solution is $$\angle UFO = \arccos\left( \frac{F_x (F_x - U_x) + F_y (F_y - U_y)}{\sqrt{\left( F_x^2 + F_y^2 \right)\left((F_x - U_x)^2 + (F_y - U_y)^2\right)}}\right)$$
I used Maple to try and see if this expression simplifies, but it does not seem to. Numerically, plugging in the above values does yield $\angle UFO = 75.000000°$.