Let $T:\ell^2\to \ell^2$ such that $T(x_1,x_2,x_3,x_4,...)=(0,4x_1,x_2,4x_3,x_4....)$
then
$T^ *:\ell^2 \to \ell^2$ such that $T^*(x_1,x_2,x_3,x_4...)=(4x_2,x_3,4x_4,....)$,
hence $(T^*)^2(x_1,x_2,x_3,x_4,...)=(4x_3,4x_4,4x_5,...)$.
I have shown that if $|\lambda|<4$, then $\lambda$ is eigenvalue of $(T^*)^2$.
Now I have to show $\sigma(T)=\{\lambda \in\mathbb{C:|\lambda|\leq2}\}$
from the fact that $|\lambda|<4$ is eigenvalue of $(T^*)^2$. We have
$$\{\lambda\in\mathbb{C:|\lambda|\leq4}\} \subseteq \sigma(T^2).$$
Now in the hint given it say if $|\lambda|\leq2$, then $\lambda\in\sigma(T)$, otherwise $\lambda^2\notin \sigma(T^2)$.
But I am not getting this hint, how does $\lambda\notin \sigma(T)$ imply $\sigma^2\notin(T^2)$?
Could anyone help me please?
What is always true is that $$\tag1\sigma(T^2)=\{\lambda^2:\ \lambda\in\sigma(T)\}=\sigma(T)^2,$$which follows easily from $T^2-\lambda^2=(T-\lambda)(T+\lambda)$. As Daniel mentioned, I don't see that you can immediately obtain $\sigma(T)$ from this.
What you get from $(1)$, since $\|T^2\|=4$, is that $$\tag2\sigma(T)\subset\{|\lambda|:\ |\lambda|\leq2\}.$$
For $T^*$, you can do a similar trick than you did for $T^{*2}$. Namely, if $|\lambda|<2$, then $$ T^*(\lambda,\frac{\lambda^2}4, \frac{\lambda^3}4,\frac{\lambda^4}{16},\frac{\lambda^5}{16},\frac{\lambda^6}{2^6},\frac{\lambda^7}{2^6}\ldots) =\lambda\, (\lambda,\frac{\lambda^2}4, \frac{\lambda^3}4,\frac{\lambda^4}{16},\frac{\lambda^5}{16},\frac{\lambda^6}{2^6},\frac{\lambda^7}{2^6}\ldots), $$ which shows that $$\tag3\{|\lambda|:\ \lambda<2\}\subset\sigma(T^*).$$ Now combining $(2)$ and $(3)$, $$ \sigma(T)=\overline{\sigma(T^*)}=\{|\lambda|:\ |\lambda|\leq2\}. $$