$A=\{1,s,r^2,sr^2\}\subseteq D_8$
We know $Z(D_8) = \{ 1,r^2\} \subseteq C_G(A) $ Hence $1,r^2 \in C_G(A) $
also $ sss^{-1} = s, \; s(sr^2)s^{-1} = sr^2 \; $ Hence $s \in C_G(A) $
Since $C_G(A)\leq G,$ using closure $sr^2 \in C_G(A)$
Hence $A\subseteq C_G(A) \Rightarrow |C_G(A)|\geq 4$
By Lagrange $|C_G(A)|=4,8 $. But $rsr^{-1} = sr^2 \neq s$ Hence $r \notin C_G(A)$
$|C_G(A)| \neq 8$. Hence $|C_G(A)|=4 $ and $C_G(A)=A$
We know $C_G(A) \leq N_G(A) \Rightarrow |N_G(A)|\geq 4$. Also $rsr^{-1}=sr^2 \in N_G(A)$ and $r(sr^2)r^{-1}=s \in N_G(A)$.
Hence $rAr^{-1} = A$. Hence $|N_G(A)|>4 \Rightarrow |N_G(A)|=8 \Rightarrow |N_G(A)|=G$
Is there a better way to solve this?
First show $A$ is a subgroup:
$s$ and $r^2$ are their own inverses. So also is $sr^2$, since it equals $r^2s$ by the group relations. Thus $A$ is closed and contains an inverse for each element. $\therefore A\le D_8$.
Then $A$ is normal in $D_8$ since $[D_8:A]=2$.
Thus $N_{D_8}(A)=D_8$.