Hyperbolic sums $S(n,k)=\sum_{m=1}^{\infty} \frac{1}{\cosh(\pi m)^{n} \sinh(\pi m)^{k}}$ It can be verified that
$$
\sum_{n=1}^{\infty}\frac{1}{\cosh(\pi n)^3\sinh(\pi n)^2}
=\frac{11}{12}-\frac{3K}{2\pi}+\frac{K^2}{2\pi^2}-\frac{K^3}{\pi^3}$$
where $K=\frac{\Gamma\left ( \frac14 \right )^2 }{4\sqrt{\pi}}$(here and below) and $\Gamma(z)$ is the Euler Gamma function.

Proof. The proof relies on complex analysis.  Let us directly consider the function
$$
f(z)=\frac{\operatorname{cs}\left(z,\frac{1}{\sqrt{2}}\right)}{\cosh(\frac{\pi z}{2K})^3\sinh(\frac{\pi z}{2K})^2}.
$$
Where $\operatorname{cs}(z,k)$ denotes one of Jacobi's elliptic functions. Integrate the function along a rectangular contour with vertices $-L,L,L+2Ki,-L+2Ki$ counterclockwise, where $L$ is a positive real number such that there is no pole on the left(or right) edge of the contour. To avoid poles on the upper and lower edges, construct infinite semicircles around the poles inside the rectangle with radius $r$.
Since $\operatorname{cs}$ has double periodicity and poles at $z=\{ (2m+2ni)K\mid (m,n)\in\mathbb{Z}^2\} $, we have
$$
\mathcal{P.V.}\int_{-L}^{L}[f(z)-f(z+2Ki)]\text{d}z+
\sum_{L_n\in\text{semicircles}}\int_{L_n}f(z)\text{d}z
+\int_{L}^{L+2Ki}f(z)\text{d}z+\int_{-L+2Ki}^{-L}f(z)\text{d}z
=2\pi i\sum_{z_k\in\text{poles inside the contour}}\operatorname{Res}(f(z),z_k).
$$
As $L\rightarrow+\infty$, the first, third and fourth integrals obviously vanish because $f(z)$ is odd. So we have
$$
\sum_{L_n\in\text{semicircles}}\int_{L_n}f(z)\text{d}z=2\pi i\sum_{z_k\in\text{poles inside the contour}}\operatorname{Res}(f(z),z_k).
$$
The rest is easy. Note that
$$
\sum_{L_n\in\text{semicircles}}\int_{L_n}f(z)\text{d}z=
-\pi i\sum_{z_k\in2K\mathbb{Z},2K\mathbb{Z}+2Ki}\operatorname{Res}(f,z_k)=-\pi i\left ( 4\sum_{n=1}^{\infty}\frac{1}{\cosh(\pi n)^3\sinh(\pi n)^2}+2\operatorname{Res}(f(z),0)\right).
$$
Then we get
$$
\sum_{n=1}^{\infty}\frac{1}{\cosh(\pi n)^3\sinh(\pi n)^2}=\frac12\left(-\operatorname{Res}(f,0)-\operatorname{Res}(f,iK)\right).
$$
The residues are easy to compute, then we prove the sum. We can compute the values of $S(2n+1,2k)=\sum_{m=1}^{\infty} \frac{1}{\cosh(\pi m)^{2n+1}
\sinh(\pi m)^{2k}},(n,k)\in\mathbb{Z}^2$ if the sum exists. For example,
$$
S(3,4)=-\frac{943}{720}+\frac{5K}{2\pi}
-\frac{13K^2}{12\pi^2}+\frac{K^3}{\pi^3}+\frac{K^4}{20\pi^4},$$
$$
\sum_{n=1}^{\infty} \frac{1}{\cosh(\pi n)^{11}}
=-\frac{1}{2}+\frac{63K^{}}{256\pi^{}}
+\frac{117469K^{3}}{201600\pi^{3}}+
\frac{17281K^{5}}{10080\pi^{5}}+\frac{3553K^{7}}{1200\pi^{7}}
+\frac{869K^{9}}{240\pi^{9}}+\frac{1381K^{11}}{600\pi^{11}}.
$$

Question 1: Can we prove the sum in an alternative way, or is there a simpler method to compute these $S(n,k)$?
Question 2: Is there a complex method to compute $S(2n,2k)$ for integers $n,k$?
 A: With regard to Question 2, the infinite series $$\sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)}$$ can actually be evaluated using the typical $\pi \cot(\pi z)$ kernel as was shown here.
However, using the kernel $\pi \cot(\pi z)$ to evaluate $$\sum_{n=1}^{\infty} \frac{1}{\cosh^{2}(\pi n)}$$ doesn't work out so nicely, and using $\operatorname{cs}\left( z, \frac{1}{\sqrt{2}}\right)$, where $\frac{1}{ \sqrt{2}}$ is the elliptic modulus, seems to result in the trivial equation $0=0$.
You might already know this, but what can be shown using contour integration is that $$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{\cosh^{2}(\pi n)} = \frac{4 K^{2}}{\sqrt{2} \pi^{2}} , $$ where $K$ is the elliptic integral of the first kind with modulus $\frac{1}{\sqrt{2}}$.
(Mathematica for some reason uses the parameter $m$ instead of the modulus $k$.)
We can exploit the fact that the Jacobi elliptic function $\operatorname{ns} \left(z, \frac{1}{\sqrt{2}} \right)$ has simple poles at $z=2\mathbb{Z} K \pm 2i \mathbb{Z} K$ with residues that alternate between $1$ and $-1$.
For anyone interested, there's a nice table here that shows special values of the Jacobi elliptic functions.
Integrating the function $$f(z) = \frac{\operatorname{ns}\left(z, \frac{1}{\sqrt{2}} \right)}{\cosh^{2} (\frac{\pi z}{2 K})} $$ around the same contour that was used in the question, we get $$-\pi i \sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(\pi n)} - \pi i\sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(\pi n)} = 2 \pi i  \operatorname{Res} \left[f(z), iK \right].$$
(The integral vanishes on the top and bottom of the contour because $f(z)$ is odd on those lines, and the integral vanishes on the left and right sides of the contour because the magnitude of $\cosh(z)$ grows exponentially as $\Re(z) \to \pm \infty$.)
Since $ \operatorname{ns} \left(z, \frac{1}{\sqrt{2}} \right)$ is analytic at $z= iK$ with value $0$ and first derivative $\frac{1}{\sqrt{2}}$ (see here), and $$\frac{1}{\cosh^{2} (\frac{\pi z}{2 K})}= - \frac{4K^{2}}{\pi^{2}} \frac{1}{(z-iK)^{2}} + \mathcal{O}(1),$$
we have $$\operatorname{Res} \left[f(z), iK \right] = - \frac{4K^{2}}{\sqrt{2} \pi^{2}}. $$
Therefore, $$ \sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(\pi n)} = \frac{4 K^{2}}{\sqrt{2} \pi^{2}} = \frac{\left(\Gamma \left(\frac{1}{4} \right)\right)^{4}}{4 \sqrt{2} \pi^{3}}.$$

This approach also shows that $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sinh^{2}(\pi n)} = \frac{1}{2} \left(\frac{K^{2}}{\pi^{2}}-\frac{1}{3} \right) = \frac{1}{2} \left(\frac{\left(\Gamma \left(\frac{1}{4} \right)\right)^{4}}{16 \pi^{3}}- \frac{1}{3} \right). $$

And by integrating the function $$\frac{\operatorname{ns} \left(z, \frac{1}{\sqrt{2}} \right)}{\cosh^{2}\left(\frac{\pi z}{K} \right)} $$ around the same contour and using the derivative $$\frac{\mathrm d}{\mathrm dz} \operatorname{ns}(z,k) = - \frac{\operatorname{cn}(z,k) \operatorname{dn}(z,k)}{\left(\operatorname{sn}(z,k)\right)^{2}}, $$ along with values from this table, we find $$\sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(2 \pi n)} = \frac{2^{1/4} (1+ \sqrt{2})K^{2}}{\pi^{2}} = \frac{2^{1/4}(1+\sqrt{2}) \left( \Gamma \left(\frac{1}{4} \right)\right)^{4}}{16 \pi^{3}}.$$
