Here is what I did for part (a):

For $1<|z|<2$, if we follow equation (12) and (13), we have

$Res(\frac{z+2}{\frac{z-2}{z+1}};-1) = \frac{z+2}{z-2}|_{z=-1}= -\frac{1}{3}$

$Res(\frac{z+2}{\frac{z+1}{z-2}};2) = \frac{z+2}{z+1}|_{z=2}= \frac{4}{3}$

By (12) and (13):

$\sum_{-\infty}^{\infty}a_nz^n$,where

$a_n = \sum\frac{1}{3}(-1)^n,n=-1,-2,...$

$and$

$a_n = -\sum\frac{4}{3}(2)^{-n-1}=-\sum\frac{4}{3}(\frac{1}{2^{n+1}}),n=0,1,2,...$

Similar for when $2<|z|<\infty$

You can read example 12 from 2.5