A couple of formulas for $\pi$
The second formula is known from Fourier analysis, and perhaps that is a place to look for the other formulas as well. The Fourier series of $f(x)={\pi-x \over 2}$ is given by $${\pi-x \over 2}=\sum_{n=1}^\infty {\sin(nx)\over n}, \qquad 0<x<2\pi$$ Let $x={2\pi\over 3}$, and you get the second formula above.
Addendum: The third formula can be shown in the same way as the second. By the method of partial fractions, the sum can be rewritten as $$4\sqrt3 \sum_{n=1}^\infty {12n-5 \over 8n(2n-1)(3n-1)(6n-5)}=3\sqrt3 \sum_{n=1}^\infty \Bigl({1\over 6n}-{1\over 6n-2}-{1\over 6n-3}+{1\over 6n-5}\Bigr)$$ The last expression is equal to $$6\sum_{n=1}^\infty {\sin\bigl((n+1){\pi\over 3}\bigr)\over n}$$ which can be summed to yield the result $\pi$.
One way is to do this is to expand the nominator $\sin((n+1){\pi\over 3})={1\over 2}\sin({n\pi\over 3})+{\sqrt3\over 3}\cos({n\pi\over 3})$ and to use the formulas $\sum_{n=1}^\infty{\sin(nx)\over n}={\pi-x\over 2}$ and $\sum_{n=1}^\infty{\cos(nx)\over n}=\log(2-2\cos(x))$, which are valid for $0<x<2\pi$, and to evaluate these expressions at $x={\pi\over 3}$.
Addendum We now derive the two remaining formulas. Partial fractions on the first formula yields $$\sum_{n=1}^\infty{3\over n(2n-1)(4n-3)}=4\sum_{n=1}^\infty\Bigl({1\over 4n}-{3\over 4n-2}+{2\over 4n-3}\Bigr)$$ Here we need more than one trigonometric function, but it is possible to rewrite this as $$4\sum_{n=1}^\infty {\sin(n{\pi\over 2})+2\cos(n{\pi\over 2})-\cos(n\pi) \over n}=4\Bigl(f\Bigl({\pi\over 2}\Bigr)+2g\Bigl({\pi\over 2}\Bigr)-g(\pi)\Bigr) = 4\Bigl({\pi\over 4}+8\log(2)-4\log(4)\Bigr)=\pi$$ where $f(x)={\pi-x\over 2}$ and $g(x)=-{1\over 2}\log(2-2\cos(x))$, as above.
Finally, partial fractions on the fourth formula yields $$16\sum_{n=1}^\infty {864n(n-1)+226\over (12n-1)(12n-5)(12n-7)(12n-11)(4n-1)(4n-3)}$$ $$=6\sum_{n=1}^\infty\Bigl(-{1\over 12n-1}+{2\over 12n-3}-{1\over 12n-5}+{1\over 12n-7}-{2\over 12n-9}+{1\over 12n-11}\Bigr) $$ $$= 6\sum_{n=1}^\infty{-{1\over 3}\sin{n\pi\over 6}+{4\over 3}\sin{3n\pi\over 6}-{1\over 3}\sin{5n\pi\over 6} \over n}$$ $$=6\Bigl(-{1\over 3}f\Bigl({\pi\over 6}\Bigr)+{4\over 3}f\Bigl({3\pi\over 6}\Bigr)-{1\over 3}f\Bigl({5\pi\over 6}\Bigr)\Bigr)=6\Bigl(-{1\over 3}{5\pi\over 12}+{4\over 3}{\pi\over 4}-{1\over 3}{\pi\over 12}\Bigr)=\pi$$ where we still have $f(x)={\pi-x\over 2}$.
Addendum: The Fourier series of a function $f(x)$ defined on the interval $[0,2\pi]$ is given by $$a_0+\sum_{n=1}^\infty\bigl(a_n\cos(nx)+b_n\sin(nx)\bigr)$$ where $$a_0={1\over 2\pi}\int_0^{2\pi}f(x){\rm d}x,\qquad a_n={1\over \pi}\int_0^{2\pi}f(x)\cos(nx){\rm d}x,\qquad b_n={1\over \pi}\int_0^{2\pi}f(x)\sin(nx){\rm d}x$$ In our case, we get $a_n=0$ for all $n\geq 0$ from the symmetry of $f(x)$ about the point $x=\pi$ (i.e., $f(\pi-x)=-f(\pi+x)$), and we get $b_n={1\over n}$ by integration by parts. The Fourier series will converge to $f(x)$ at all points where (the periodic extension of) $f$ is differentiable.
The first formula may be rewritten by changing the lower limit of the summation:
$$\begin{align} \frac{\pi}{3} &=\sum_{n=1}^\infty \frac{1}{n(2n-1)(4n-3)} \\ &=\sum_{n=0}^\infty \frac{1}{(n+1)(2(n+1)-1)(4(n+1)-3)} \\ &=\sum_{n=0}^\infty \frac{1}{(n+1)(2n+1)(4n+1)} \\ \end{align}$$
This last equation is given by Lehmer in page 139 here.
A similar one that uses four factors is $$\pi-3=\sum_{k=0}^\infty \frac{24}{(4k+2)(4k+3)(4k+5)(4k+6)}$$
(see Series and integrals for inequalities and approximations to $\pi$)
A formula which is similar to the third one you give is: $$\pi=72 \sqrt{3} \sum_{k=0}^\infty \frac{1}{(6 k+1) (6 k+2) (6 k+4) (6 k+5)}$$ which has all terms positive and constant numerator. Taking the first term of the series out of the summation, this proves $\pi>\frac{9}{5}\sqrt{3}\approx 3.12$.
There is also $$\pi=\frac{23040(17+9 \sqrt{3})}{23} \sum_{k=0}^\infty \frac{1}{(12 k+1) (12 k+3) (12 k+5) (12 k+7) (12 k+9) (12 k+11)}$$ with the denominators of your fourth formula. The first term in this series leads to the approximation
$$\pi \approx \frac{2^9(17+9\sqrt{3})}{5313}=\frac{2^{10}((2^4+1)+(2^3+1)\sqrt{3})}{21·22·23} = \frac{2^{10}}{231(17-9\sqrt{3})}\approx 3.140$$
Using $163$ instead of $\frac{5313}{17+9\sqrt{3}} \approx 163.033$ gets one more correct digit. The approximation $$\pi \approx \frac{2^9}{163} \approx 3.1411$$ is due to Stoschek (see http://mathworld.wolfram.com/PiApproximations.html)
Also compare to $\pi \approx \frac{2^{13}}{3(-383+560\sqrt{5})}$ given by T.Piezas here.