c. $1-\cos z\over z^2$ and d. $\frac 1{(z-\pi)^4}$.
c. This is what I have got so far
removable singular point at $z_0=0$ of the function, $f(0)=1/2$, $m=1$ and $b=1/2$
residue =$1/2$ as series =$1/2!-z^4/4!+z^4/6!....$
d. $\frac 1{(z-\pi)^4}$ still confused on this one. Would it be an essential singularity =0 for z=pi ??? Series (1/z)^4+4pi/z^5+10pi^2/z^6+....?????
If a singularity is removable then the residue is $0$. This should be obvious from the definition of a residue as the coefficient in front of $\frac 1{z-z_0}$ in the series expansion.
For the second one, your series expansion is wrong. Find the correct one and you will have the answer at once. Hint: it is MUCH easier than what you wrote.
Another hint: if a meromorphic function has an antiderivative in a neighbourhood of a pole, then automatically its residue at that pole is $0$. What makes $1\over z-z_0$ so special is that it doesn't have an antiderivative around $z_0$, unlike any other power (positive or negative) of $z-z_0$.