$$\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}=\sum^\infty_{m=0}\sum^{10}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$
As the sum at \(n=5\) and \(n=10\) are positive and negative infinite sum, they cancel and we are left with,
$$\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}=\sum^\infty_{m=0}\sum^{2,3,4,6,7,8,9}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$
We have eventually,
$$\cfrac{\pi^2}{25}\left[\sum^{2,4,5}_{i=1}csc^2(\cfrac{(2i-1)\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+(2i-1))^3} +\\ \cfrac{1}{8}\sum^{4}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}+\\ \sum^{2,3,4,6,7,8,9}_{n=1}{\cfrac{1}{n^3}}csc^2(\cfrac{n\pi}{5})\right]$$
This will converge and can be solved, numerically.
