Let $f(z):=\dfrac{1}{2-z-z^2}, z\in\mathbb{C}\setminus\left\{ {1, -2}\right\}$.
i) Express $f$ in the form $\dfrac{A}{1-z}+\dfrac{B}{2+z}$. [Answer to this is $\dfrac{1/3}{1-z}+\dfrac{1/3}{2+z}$].
ii) Write down the Taylor expansion of $f$ on the disk $|z|<1$.
iii) Write down the Laurent expansion of $f$ on the annulus $1<|z|<2$.
Not sure of what to do here. I assume Laurent's Theorem is useful.
I'll solve $\text{(iii)}$ after which you should be able to handle $\text{(ii)}$.
Let $z\in \mathbb C\setminus \{-2, 0, 1\}$.
The following holds:
$$\dfrac{1}{1-z}=-\dfrac 1 z\dfrac{1}{1-\frac 1 z}=-\dfrac 1 z\sum \limits _{n=0}^\infty\left(\left(\dfrac 1 z\right)^n\right)= \sum \limits _{n=0}^\infty\left(-\left(\dfrac 1 z\right)^{n+1}\right), \text{ if }\underbrace{\left|\dfrac 1 z\right|<1}_{\iff \large{1<|z|}},$$
$$\dfrac 1{2+z}=\dfrac 1 2\dfrac 1{1-\left(-\frac z 2\right)}=\dfrac 1 2\sum \limits _{n=0}^\infty\left(\left(-\dfrac z 2\right)^n\right)=\sum \limits _{n=0}^\infty\left(\dfrac {(-1)^n}{2^{n+1}}z^n\right), \text{ if }\underbrace{\left|\dfrac z 2\right|<1}_{\iff \large{|z|<2}}.$$
Factoring in $\dfrac 1 3$ and adding you get the laurent series in the set $\{z\in \mathbb C\colon 1<|z|\}\cap \{z\in \mathbb C\colon |z|<2\}$, that is, you get the laurent series in the annulus $1<|z|<2$.