Calculate the first return probability for the following states in the matrix.
Where the states are 1,2 and 3.
for $f_{33}^{1}=p_33=0$
for $f_{33}^{2}=p_{32} p_{23}=(1)*(1/2)$
for $f_{33}^{3}=p_{32}p_{23}p_{33}=0$
for $f_{33}^{4}=p_{32}p_{23}p_{32}p_{23}=1*(1/2)(1)(1/2)=1/4$
So I get $f_{33}^{2n}=(1/2)^n$
and $f_{33}^{2n-1}=0$
But I am not sure how to find $f_{11} $ and $f_{22}$
Assuming that the matrix that you presented is $P^T$, then let us start with $f_{22}$. Once the chain is in state $2$ it either goes (w.p. $1/2$) to $1$ or $3$ and then w.p. $1$ goes back to $2$, hence, $$ f_{22}(2) = \frac{1}{2}\times 1 + \frac{1}{2}\times 1 = 1, $$ and $0$ for any other $n$. As for $f_{11}$ - note that this is a periodic chain with period $2$, hence the chain can return to $1$ with any number of $2n$ steps, in particular $$ f_{22}(2n) = \frac{1}{2^n}\,. $$ I.e., the chain goes to state number $2$ in the first step and then alternates $n\ge 0$ times between $2$ and $3$ until it goes back from $2$ to $1$.