Let $$h(z)=\frac{1}{\sin(4z-\pi)}$$ What kind of singularity does function $h(z)$ at the point $z=\frac{\pi}{4}$?
I know from theory that the point is essential singularity if and only if $\lim_{z \to z_0}h(z)$ does not exist OR Laurent series $a_{-m} \ne 0 $ $\forall m\in\Bbb{N}$.
First of all, I don't know, how to calculate the limit by myself without using wolfram alpha and to see that the limit does not exit so $z=\frac{\pi}{4}$ is essential singularity. On the other hand, wolfram alpha says that Laurent series of function h(z) has $a_{-1}$ and that contradicts $a_{-m} \ne 0 $ $\forall m\in\Bbb{N}$.
So, my question is: how to solve this problem and why two different methods give me different answers? Or maybe this is not even essential singularity?
Your theoretical part is not correct. The point $\frac\pi4$ would be an essential singularity if and only if $a_m\neq0$ for infinitely many $n$'s, not for all of them.
Anyway, $\lim_{z\to\frac\pi4}\frac{z-\frac\pi4}{\sin(4z-\pi)}=\frac14$ and therefore your function has a pole of order $1$ at $\frac\pi4$.