Computing alternating sum using contour integration

In general, if $k$ is a positive integer greater than or equal to $2$, $$\sum_{m=1}^{\infty} \frac{(-1)^{m-1} \sin^{k}(\alpha m)}{ m^{k}} = \frac{\alpha^{k}}{2} \, , \quad |\alpha| \le \frac{\pi}{k}.$$

We first need to argue that $$\lim_{N \to \infty} \int_{|z|=N+\frac{1}{2}} \left(\frac{\sin\alpha z}{ z}\right)^k \frac{\pi}{\sin \pi z} \, dz = 0$$ if $|\alpha| \le \frac{\pi}{k}$.

But this follows from the fact that as $\text{Im}(z) \to \pm \infty$, $\left|\frac{\sin^{k}(\alpha z)}{\sin(\pi z)} \right|$ behaves like $\frac{1}{2^{k-1}} e^{\pm (k |\alpha|-\pi) \, \text{Im}(z)}$.

So we have

$$ \begin{align} 2\sum_{m=1}^{\infty} \frac{(-1)^{m-1} \sin^{k}(\alpha m)}{ m^{k}} &= \text{Res} \left[\left(\frac{\sin\alpha z}{ z}\right)^k \frac{z}{\sin \pi z} , \, 0 \right] \\ &= \lim_{z \to 0} \left(\frac{\sin\alpha z}{ z}\right)^k \frac{\pi z}{\sin \pi z} \\ &=a^{k}(1) \\&= a^{k}. \end{align}$$

You calculated residues wrongly. By your series around $z=0$ you can only find residue at $z=0$. When $z=n$ you know that poles are of first order, so use the formula

$$R_a=\lim_{z\to a} \frac{\pi(z-a)}{\sin(\pi z)}$$

You will get that $R_0=1,R_1=-1,R_2=1,...$ and your formula will look like the following.

$$2\pi i \left(2 \sum\limits_{n=1}^\infty(-1)^n \frac{\sin ^2 n\alpha}{(n \alpha)^2} + 1\right) = 0$$