Compute polylog of order $3$ at $\frac{1}{2}$
Note Landen's identity
$$\color{red}{{\rm Li}_2(z)+{\rm Li}_2\left(\frac{z}{z-1}\right)=-\frac{1}{2}\ln^2(1-z)}\tag1$$
which can be easily derived by differentiating ${\rm Li}_2\left(\frac{z}{z-1}\right)$ then integrating back. Plucking in $z=.5$ yields
$$\color{red}{{\rm Li}_2\left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}}$$
since $\displaystyle{\rm Li}_2(-1)=-\eta(1)=\small{-\frac{\pi^2}{12}}$. We will use the same idea to evaluate ${\rm Li}_3\left(\frac{1}{2}\right)$.
Observe $D_z{\rm Li}_{s+1}\left(\frac{z}{z-1}\right)=\frac{1}{z(1-z)}{\rm Li}_{s}\left(\frac{z}{z-1}\right)$. So dividing $(1)$ throughout by $z(1-z)$ then integrating gives us
$${\rm Li}_3(z)+\int\frac{{\rm Li}_2(z)}{1-z}{\rm d}z+{\rm Li}_3\left(\frac{z}{z-1}\right)=\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\int\frac{\ln^2(1-z)}{z}{\rm d}z\tag2$$
The integral on the LHS evaluates to
\begin{align}
\int\frac{{\rm Li}_2(z)}{1-z}{\rm d}z
&=-{\rm Li}_2(z)\ln(1-z)-\int\frac{\ln^2(1-z)}{z}{\rm d}z
\end{align}
so we can rewrite $(2)$ as
\begin{align}
{\rm Li}_3(z)+{\rm Li}_3\left(\frac{z}{z-1}\right)={\rm Li}_2(z)\ln(1-z)+\frac{1}{6}\ln^3(1-z)+\frac{1}{2}\int\frac{\ln^2(1-z)}{z}{\rm d}z\tag3
\end{align}
and the integral on the RHS of $(3)$ is
\begin{align}
\frac{1}{2}\int\frac{\ln^2(1-z)}{z}{\rm d}z
&=\frac{1}{2}\ln{z}\ln^2(1-z)+\int\frac{\ln{z}\ln(1-z)}{1-z}{\rm d}z\\
&=\frac{1}{2}\ln{z}\ln^2(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\int\frac{{\rm Li}_2(1-z)}{1-z}{\rm d}z\\
&=\frac{1}{2}\ln{z}\ln^2(1-z)+{\rm Li}_2(1-z)\ln(1-z)-{\rm Li}_3(1-z)+C\tag4\\
\end{align}
Plucking $(4)$ into $(3)$ then moving all the trilog terms to one side gives us
\begin{align}
&{\rm Li}_3(z)+{\rm Li}_3(1-z)+{\rm Li}_3\left(\frac{z}{z-1}\right)\\
=&\frac{1}{6}\ln^3(1-z)+\Big{[}{\rm Li}_2(z)+{\rm Li}_2(1-z)\Big{]}\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+C\\
=&\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\ln{z}\ln^2(1-z)+\frac{\pi^2}{6}\ln(1-z)+C\tag5
\end{align}
where we applied the dilogarithm reflection formula
$$\color{red}{{\rm Li}_2(z)+{\rm Li}_2(1-z)=\frac{\pi^2}{6}-\ln{z}\ln(1-z)}\tag6$$
to get to $(5)$. This formula can be easily derived by differentiating ${\rm Li}_2(1-z)$ then integrating back. Now, by letting $z=0$ in $(5)$, it is evident that $C={\rm Li}_3(1)=\zeta(3)$. Therefore,
\begin{align}&\color{red}{{\rm Li}_3(z)+{\rm Li}_3(1-z)+{\rm Li}_3\left(\frac{z}{z-1}\right)}\\=&\color{red}{\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\ln{z}\ln^2(1-z)+\frac{\pi^2}{6}\ln(1-z)+\zeta(3)}\tag7\end{align}
Finally, letting $z=.5$ in $(7)$, we get
$$2{\rm Li}_3\left(\frac{1}{2}\right)-\frac{3}{4}\zeta(3)=-\frac{1}{6}\ln^3{2}+\frac{1}{2}\ln^3{2}-\frac{\pi^2}{6}\ln{2}+\zeta(3)$$
which implies
$$\color{blue}{{\rm Li}_3\left(\frac{1}{2}\right)=\frac{7}{8}\zeta(3)-\frac{\pi^2}{12}\ln{2}+\frac{1}{6}\ln^3{2}}\tag8$$
$$\begin{aligned}I_1&=\int_0^{1/2}\frac{\ln^2(1-x)}x\mathrm {d}x\\ I_2&=\int_0^{1/2}\frac{\ln x\ln(1-x)}x\mathrm {d}x\\ I_3&=\int_0^{1/2}\frac{\ln^2 x}{1-x}\mathrm {d}x\\ I_4&=\int_0^{1/2}\frac{\ln x\ln(1-x)}{1-x}\mathrm {d}x\end{aligned}$$ $$I_1+I_3=\int_0^1\frac{\ln^2 x}{1-x}\mathrm {d}x=2\zeta(3)$$ $$I_2+I_4=\int_0^1\frac{\ln x\ln(1-x)}x\mathrm {d}x=\zeta(3)$$ $$I_1=\ln^2(1-x)\ln x\big|_0^{1/2}+2\int_0^{1/2}\frac{\ln(1-x)\ln x}{1-x}\mathrm{d}x=-\ln^32+2I_4$$ $$I_2-I_1=\int_0^{1/2}\ln\frac x{1-x}\frac{\ln(1-x)}x\mathrm{d}x=-\int_0^1\frac{\ln u\ln(1+u)}{u(1+u)}\mathrm{d}u=\frac58\zeta(3)$$(using substitution $u=\frac{x}{1-x}$)
Hence,$$I_1=\frac14\zeta(3)-\frac13\ln^32,\ I_2=\frac78\zeta(3)-\frac13\ln^32,$$ $$I_3=\frac74\zeta(3)+\frac13\ln^32,\ I_4=\frac18\zeta(3)+\frac13\ln^32.$$
$$\operatorname{Li}_3\left(\frac12\right)=\int_0^{1/2}\frac{\operatorname{Li}_2(x)}x\mathrm{d}x\\ =\operatorname{Li}_2(x)\ln x\Big|_0^{1/2}+\int_0^{1/2}\frac{\ln(1-x)\ln x}x\mathrm{d}x\\ =-\frac1{12}\pi^2\ln2+\frac12\ln^32+\frac78\zeta(3)-\frac13\ln^32$$