$\DeclareMathOperator{\ran}{im}$
If $T$ is normal, $\ker T^2 = \ker T$ and $\ran T^2 = \ran T$.
Clearly $\ker T ⊂ \ker T^2$ since $T(v) = 0 \implies T^2(v) = 0$.

Now if $T^2 v = 0$ for some $v$, then $0 = ⟨T^2 v, v⟩ = ⟨Tv, T^* v⟩$. Since $T$ is normal, $\ker T=\ker T^*$. Therefore $Tv = 0 \iff T^* v = 0$. Clearly then they are both zero, so $v ∈ \ker T$. That is $\ker T^2 ⊂ \ker T$ and we are done.

For $\ran T^2$, we use
\[\ran T = \left(\ker T^*\right)^⟂ = \left(\ker T\right)^⟂ = \left(\ker T^2\right)^⟂ = \ran (T^2)^* = \ran T^2\]
since $\ran S = \ran S^*$ for normal operators $S$. ($T^2$ is normal since $TT(TT)^* = (TT)^*(TT)$ which is easily confirmed by using $TT^* = T^*T$.)