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I numerically discovered the following conjecture:
$$Re,operatorname{Li}_2left(frac12+frac i6right)stackrel{color{gray}?}=frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).$$
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 $Im,operatorname{Li}_2left(frac12+frac i6right)$?

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 $p,qinmathbb Q$?

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:

$$
ln(frac{1}{2}pm frac{i}{6})=ln{(sqrt{frac{1}{2^2}+frac{1}{6^2}}{e}^{pm iarctan{frac{1}{3}}})}\
=ln{(sqrt{frac{5}{18}}{e}^{pm iarctan{frac{1}{3}}})}\
=frac{1}{2}ln{(frac{5}{18})}pm iarctan{frac{1}{3}}
$$

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:

$$
Re{{Li}_{2}(frac{1}{2}+iq)}=frac{{pi}^{2}}{12}-frac{1}{8}{ln{(frac{1+4q^2}{4})}}^{2}-frac{{arctan{(2q)}}^{2}}{2}
$$

First of all we know that:

$$
operatorname{Li}_2(z) = -operatorname{Li}_2left(frac{z}{z-1}right)-frac{1}{2}ln^2(1-z), quad z notin (1,infty).tag{$diamondsuit$}
$$

Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:

$$operatorname{Li}_2left(e^{itheta}right) = operatorname{Sl}_2(theta)+ioperatorname{Cl}_2(theta), quad theta in [0,2pi).tag{$heartsuit$}$$

where $operatorname{Cl}_2$ and $operatorname{Sl}_2$ are the standard Clausen functions, defined as:
$$begin{align}
operatorname{Cl}_2(theta) &= sum_{k=1}^{infty}frac{sin(ktheta)}{k^2}, \
operatorname{Sl}_2(theta) &= sum_{k=1}^{infty}frac{cos(ktheta)}{k^2}.
end{align}$$
Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that

$$
operatorname{Sl}_2(theta) = frac{pi^2}{6}-frac{pitheta}{2}+frac{theta^2}{4}, quad theta in [0,2pi).tag{$spadesuit$}
$$

Now let $z:=tfrac{1}{2}+tfrac{i}{6}$. Because $left|tfrac{z}{z-1}right| = 1$, the equation
$$
e^{itheta} = frac{z}{z-1} = -frac{4}{5}-frac{3}{5}i,
$$
has the only solution $theta = arctanleft(tfrac{3}{4}right) + pi$ in $[0,2pi)$.

Because of $(diamondsuit)$ and $(heartsuit)$ we have
$$
operatorname{Li}_2(z) = -color{red}{operatorname{Sl}_2(theta)} - i color{green}{operatorname{Cl}_2(theta)} - color{blue}{frac{1}{2}ln^2(1-z)},
$$
for $z=tfrac{1}{2}+tfrac{i}{6}$ and $theta = arctanleft(tfrac{3}{4}right) + pi$.

For the logarithm term we get
$$
Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{8}left(ln^2left(frac{18}{5}right)-(pi-2arctan 3)^2right)
$$
and
$$
Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
$$

We know that $color{red}{operatorname{Sl}_2(theta)}$ and $color{green}{operatorname{Cl}_2(theta)}$ are real quantities. By using $(spadesuit)$ for the SL-type Clausen term we get
$$
color{red}{operatorname{Sl}_2(theta)} = frac{pi^2}{12}-frac{1}{4}arctan^2left(frac{3}{4}right).
$$

Now we could obtain your conjectured closed-form:
$$
Releft[operatorname{Li}_2(z)right] = -color{red}{operatorname{Sl}_2(theta)} - Re{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} = frac{7pi^2}{48}-frac{arctan^22}3-frac{arctan^23}6-frac18ln^2!left(frac{18}5right).
$$

For the imaginary part we have
$$begin{align}
Imleft[operatorname{Li}_2(z)right] &= -color{green}{operatorname{Cl}_2(theta)} - Im{left[color{blue}{frac{1}{2}ln^2(1-z)}right]} \ &= -operatorname{Cl}_2left(arctanleft(frac{3}{4}right)+piright)-frac{1}{4}lnleft(frac{18}{5}right)(pi-2arctan 3).
end{align}$$

By using $(diamondsuit), (heartsuit)$ and $(spadesuit)$, you could generalize this process for all $z in mathbb{C}$ such that $left|frac{z}{z-1}right|=1$.

This is an answer finding $Reoperatorname{Li}_2left(frac{1+ti}2right)$ via integration method. (Personally I don't like the reflection formula)

First, we restrict $tinmathbb R$. One can prove $Reoperatorname{Li}_2left(frac{1+ti}2right)$ is differentiable and the following changing the positions of signs is correct.
$$begin{aligned}
Reoperatorname{Li}_2left(frac{1+ti}2right)&=-frac12int_0^1lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}x\
&=-frac12intint_0^1frac{partial}{partial t}lnleft(1-x+frac{1+t^2}4x^2right)frac{d x}xd t\
&=-intint_0^1frac{tx}{4-4x+(1+t^2)x^2}d xd t\
&=-intleft(frac1{1+t^2}arctan t+frac12cdotfrac t{1+t^2}lnfrac{1+t^2}4right)d t\
&=-frac12arctan^2t-frac18ln^2frac{1+t^2}4+C
end{aligned}$$
Substitute $t=0$ into the equation, we get $frac1{12}pi^2-frac12ln^22=0-frac18ln^24+C$, or $C=frac1{12}pi^2$.
Hence $$Reoperatorname{Li}_2left(frac{1+ti}2right)=frac1{12}pi^2-frac12arctan^2t-frac18ln^2frac{1+t^2}4$$