Determine the character of the singularity at $z=-2$ for function$\frac{1}{(z+2)^{2} \sin z}.$ Determine the character of the singularity at $z=-2$ for function
$$\frac{1}{(z+2)^{2} \sin z }.$$
Since function $ z \mapsto \frac{1}{\sin z} $ is holomorphic in some neighbourhood around $z=-2$, its development to Laurent's series equals to develompent to Taylor's series. Using that development I could read what character is singularity $z=-2$, but I got stuck right after first step ($\sin z $ has known development to Taylor's so I have put that in the denominator). Is that good way of solving this? I am begginer in this area, so I am not sure what exactly I am "allowed" to do in $\mathbb{C}$. Any hint helps!

What you say is correct.
The function $1/\sin z$ is analytic in a suitable neighborhood of $z = -2$. Therefore, in such a neighborhood we have:
\begin{align}
f(z) & = \frac{1}{(z+2)^{2}}\frac{1}{\sin z} \\ & = \frac{1}{(z+2)^{2}} \left\{ a_0 +a_1 (z+2)+a_2 (z+2)^{2}+ a_3 (z+2)^{3} + \cdots \right\} \\ & = \frac{a_0}{(z+2)^2} + \frac{a_1}{z+2} + a_2 + a_3 (z+2) + \cdots
\end{align}
Notice that $ a_{0} = 1/\sin(-2) \neq 0$, which implies that $z =-2$ is a pole of order 2.
