You can evaluate the alternating hyperbolic series $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{2n-1} \sinh (\pi k)}$$ for any positive even value of $n$ by integrating the function $$\frac{\pi \csc (\pi z)}{z^{2n-1} \sinh (\pi z)}$$ around a square contour that avoids the poles on both the real and imaginary axes.
For example, $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{3} \sinh (\pi k)} = - \frac{\pi^3}{360} $$ and $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{7} \sinh (\pi k)} = - \frac{13 \pi^7}{453,600}. $$
However, if you try that same thing for a positive odd value of $n$, the residues inside of the contour cancel each other, and you're left with the trivial equation $0=0$.
So what would be a way to show that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k \sinh (\pi k)} = \frac{\pi}{12} - \frac{\log 2}{2} \, ?$$
Furthermore, do closed form expressions for $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{5} \sinh (\pi k)} $, $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{9} \sinh (\pi k)} $, etc. exist?
It seems like the best way to evaluate the case $n=1$ is to express the series in terms of the lattice sum $$\sum_{i=-\infty}^{\infty} \sum_{j=-\infty}^{\infty} \frac{(-1)^{i+j}}{i^{2}+j^{2}}, \quad (i,j) \ne (0,0), $$ as explained here, and then use the Lambert series $$\theta_{4}^{2}(0,q) = 1 + 4 \sum_{m=1}^{\infty} \frac{(-1)^{m}q^{m}}{1+q^{2m}} $$ to evaluate the lattice sum (as explained here).
