Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$ I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).$$
It holds numerically with a precision of more than $30000$ decimal digits. 
Could you suggest any ideas how to prove it?
Can we find a closed form for $\Im\,\operatorname{Li}_2\left(\frac12+\frac i6\right)$?
Is there a general method to find closed forms of expressions of the form $\Re\,\operatorname{Li}_2(p+iq)$, $\Im\,\operatorname{Li}_2(p+iq)$ for $p,q\in\mathbb Q$?

I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.
We know that :
$$
{Li}_{2}(\bar{z})=\bar{{Li}_{2}(z)}
$$
So :
$$
\Re{{Li}_{2}(z)}=\frac{\bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
$$
So:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(\frac{1}{2}-\frac{i}{6})}{2}\\
=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(1-(\frac{1}{2}+\frac{i}{6}))}{2}
$$
Now let's use:
$$
{Li}_{2}(z)+{Li}_{2}(1-z)=\frac{{\pi}^{2}}{6}-\ln{z}\ln{(1-z)}
$$
So we get:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(1-(\frac{1}{2}+\frac{i}{6}))}}{2}\\
=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}}{2}
$$
Let's compute those logarithms:
$$
\ln(\frac{1}{2}\pm \frac{i}{6})=\ln{(\sqrt{\frac{1}{2^2}+\frac{1}{6^2}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\
=\ln{(\sqrt{\frac{5}{18}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\
=\frac{1}{2}\ln{(\frac{5}{18})}\pm i\arctan{\frac{1}{3}}
$$
Taking their product:
$$
\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}=\frac{1}{4}{\ln{(\frac{5}{18}})}^{2}+{(\arctan{\frac{1}{3}})}^{2}
$$
Finally:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{18}{5}})}^{2}-\frac{1}{2}{(\arctan{1/3})}^{2}\\
=\frac{7(\pi)^2}{48}-\frac{{(\arctan{2})}^{2}}{3}-\frac{{(\arctan{3})}^{2}}{6}-\frac{1}{8} {\ln{\frac{18}{5}}}^{2}
$$
There's a general formula using the same method. But the imaginary part doesn't have a known closed form:
$$
\Re{{Li}_{2}(\frac{1}{2}+iq)}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{1+4q^2}{4})}}^{2}-\frac{{\arctan{(2q)}}^{2}}{2}
$$
