Which are the poles of this integral and what order do they have?
$$\oint_D\left(\frac{\cos(\pi z/2)}{(z^2+i)(\sin^{2}(\pi z)}\right)\,dz$$
where $D$ is $|z|<3/2$.
In my opinion the should be $z= -\frac{\sqrt(2)}{2}+i\frac{\sqrt(2)}{2}$,
$z= \frac{\sqrt(2)}{2}-i\frac{\sqrt(2)}{2}$
poles of the first order.
and $z=0$ pole of the second order
and $z=1$ pole of the first order.
But wolfram doesn't find those two. do you know which are the poles for this function ?
It is probably best to write the function so that you do not have shared zeroes of numerator and denominator. We have $\sin(\pi z )=2\sin \frac{\pi z}2 \cos\frac{\pi z}2$ so your function is $$f(z) = \frac{1}{4(z^2+i)\sin^2\frac{\pi z}2\cos \frac{\pi z }2} $$ This clearly has a simple pole at the zeroes of $z^2+i$ that you calculated correctly above, simple poles at the zeroes of $\cos \frac{\pi z}2$, so at $+1$ and $-1$, and double poles at the zeroes of $\sin \frac{\pi z}2$. The only zero of that function in $D$ is $0$.
So summing up, all you have missed is an additional simple pole at $-1$.