Reduction of dilogarithm. Known identities for the dilogarithm allow us the following simple transform:
$$ \operatorname{Li}_2\left(\tfrac{1}{2} + \tfrac{i}{2} \tan\theta\right) = - \operatorname{Li}_2(e^{i(\pi+2\theta)}) -\tfrac{1}{2}\log^2\left(\tfrac{1}{2} - \tfrac{i}{2} \tan\theta \right). $$
In this case, we have $\theta = \tfrac{1}{12}\pi$ and hence the dilogarithm in question is written as
$$ \operatorname{Li}_2\left(\tfrac{1}{2} + \tfrac{i}{2}\tan\tfrac{\pi}{12} \right) = - \operatorname{Li}_2(e^{7\pi i/6}) -\tfrac{1}{2}\log^2\left( \tfrac{1}{2} - \tfrac{i}{2}\tan\tfrac{\pi}{12} \right). \tag{1} $$
Now utilizing the Fourier series of the Bernoulli polynomial, we know that $\operatorname{Li}_2(e^{i\theta})$ reduces to
$$
\operatorname{Li}_2(e^{i\theta})
= \sum_{k=1}^{\infty} \tfrac{1}{k^2}(\cos (k\theta) + i\sin(k\theta))
= \tfrac{1}{6}\pi^2 - \tfrac{1}{2}\pi\theta + \tfrac{1}{4}\theta^2 + i \operatorname{Cl}_2(\theta) \tag{2}
$$
for $0 \leq \theta \leq 2\pi$, where $\operatorname{Cl}_2(\theta) = \sum_{k=1}^{\infty} k^{-2}\sin(k\theta)$ is the Clausen function. So it remains to simplify $\operatorname{Cl}_2 \left( \tfrac{7}{6}\pi \right)$.
Reduction of Clausen function. To this end, we group the terms
$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = \sum_{k=1}^{\infty} \frac{1}{k^2} \sin \left(\tfrac{7}{6}k\pi\right) $$
according the value of sine and simplifying each group as in this proof, we find that
$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right)
= \tfrac{1}{144} \sum_{j=1}^{6} \sin \left(\tfrac{7}{6}j\pi\right) \left( \psi^{(1)}\left(\tfrac{j}{12}\right) - \psi^{(1)}\left(\tfrac{1}{2}+\tfrac{j}{12}\right) \right). \tag{3} $$
Now utilizing the reflection formula and the duplication and triplication formula extensively, we can simplify the above sum as
$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = -\tfrac{2}{3}G - \tfrac{1}{12\sqrt{3}}\pi^2 + \tfrac{1}{8\sqrt{3}} \psi ^{(1)}\left(\tfrac{1}{3}\right). \tag{4} $$
Indeed, we expand the summation in (3) and utilize the triplication formula to the green-colored groups and the duplication formula to the blue-colored groups to obtain
\begin{align*}
\operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right)
&
= - \tfrac{1}{288} \psi^{(1)}\left(\tfrac{1}{12}\right)
+ \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{6}\right)
- \tfrac{1}{144} \psi^{(1)}\left(\tfrac{1}{4}\right)
+ \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{3}\right)
- \tfrac{1}{288} \psi^{(1)}\left(\tfrac{5}{12}\right) \\
&\quad + \tfrac{1}{288} \psi^{(1)}\left(\tfrac{7}{12}\right)
- \tfrac{1}{96\sqrt{3}} \psi^{(1)}\left(\tfrac{2}{3}\right)
+ \tfrac{1}{144} \psi^{(1)}\left(\tfrac{3}{4}\right)
- \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{5}{6}\right)
+ \tfrac{1}{288} \psi^{(1)}\left(\tfrac{11}{12}\right) \\
&
= - \tfrac{1}{288} \color{green}{\left( \psi^{(1)}\left(\tfrac{1}{12}\right)
+ \psi^{(1)}\left(\tfrac{5}{12}\right)
+ \psi^{(1)}\left(\tfrac{9}{12}\right) \right)}
- \tfrac{3}{288} \psi^{(1)}\left(\tfrac{1}{4}\right) \\
&\quad
+ \tfrac{1}{288} \color{green}{\left( \psi^{(1)}\left(\tfrac{3}{12}\right)
+ \psi^{(1)}\left(\tfrac{7}{12}\right)
+ \psi^{(1)}\left(\tfrac{11}{12}\right) \right)}
+ \tfrac{3}{288} \psi^{(1)}\left(\tfrac{3}{4}\right) \\
&\quad
+ \tfrac{1}{96 \sqrt{3}} \color{blue}{\left( \psi^{(1)}\left(\tfrac{1}{6}\right)
+ \psi^{(1)}\left(\tfrac{2}{3}\right) \right)}
+ \tfrac{1}{48 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{3}\right) \\
&\quad
- \tfrac{1}{96 \sqrt{3}} \color{blue}{\left( \psi^{(1)}\left(\tfrac{1}{3}\right)
+ \psi^{(1)}\left(\tfrac{5}{6}\right) \right)}
- \tfrac{1}{48\sqrt{3}} \psi^{(1)}\left(\tfrac{2}{3}\right) \\
&
= \tfrac{1}{16 \sqrt{3}} \left( \psi^{(1)}\left(\tfrac{1}{3}\right) - \psi^{(1)}\left(\tfrac{2}{3}\right) \right)
- \tfrac{1}{24} \left( \psi^{(1)}\left(\tfrac{1}{4}\right) - \psi^{(1)}\left(\tfrac{3}{4}\right) \right).
\end{align*}
Now we focus on the last line. Applying the reflection formula to the first term and comparing the definition of the Catalan constant $G$ with the second term, we obtain (4) as claimed.
Finally, plugging this back gives the desired result.
Addendum: Fourier series of the Bernoulli polynomial. Taking imaginary part of
$$ \log(1-e^{2\pi i x}) = - \sum_{k=1}^{\infty} \frac{e^{2\pi i k x}}{k} $$
for $x \in (0, 1)$, we find that the Bernoulli polynomial $B_1(x)$ of degree 1 is written as
$$ B_1( x ) = x - \tfrac{1}{2} = - \frac{1}{2\pi i} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k}, \quad 0 < x < 1.$$
Integrating both sides repeatedly and using the relation $B_n'(x) = nB_{n-1}(x)$, we find that for any $n \geq 1$,
$$ B_n( x ) = - \frac{n!}{(2\pi i)^n} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k^n}, \quad 0 < x < 1.$$
Notice that depending on whether $n$ is even or odd, this reduces to either cosine series or sine series. For example, when $n = 2$ we have
$$ x^2 - x + \tfrac{1}{6} = B_2(x) = \frac{1}{\pi^2} \sum_{k=1}^{\infty} \frac{\cos(2\pi k x)}{k^2} $$
and hence we obtain the formula which was used in our solution:
$$ \sum_{k=1}^{\infty} \frac{\cos(k x)}{k^2} = \pi^2 B_2\left(\tfrac{1}{2\pi} x\right) = \tfrac{1}{4} x^2 - \tfrac{1}{2}\pi x + \tfrac{1}{6}\pi^2. $$