please give me a hand.
let $ a \in A $ be a normal element of the C$ ^{*}$- algebra $ A$.
a) show that $ \sigma ( a ) = \{ 0 , 1 \} $ if only if $a $ is a projection ($ a^{2} = a = a^{*}$).
b) if $a$ is hermitian ($a =a^{*}$) and $ a^{3} = a^{2}$, then $a$ is projection.
If $a$ is a projection, then $a^2-a=0$. That is $p(a)=0$, where $p$ is the polynomial $p(t)=t^2-t$. By the Spectral Mapping Theorem, $$\{0\}=\sigma(p(a))=p(\sigma(a)).$$ Then $\lambda^2-\lambda=0$ for all $\lambda\in\sigma(a)$. This shows that $\sigma(a)\subset\{0,1\}$. The equality does not necessarily hold: for instance $\sigma(0)=\{0\}$, and $\sigma(I)=\{1\}$.
If $\sigma(a)=\{0,1\}$, one can consider the abelian C$^*$-subalgebra $A_0$ generated by $a$. We have $$\{0,1\}=\sigma(a)=\{\omega(a):\ \omega\in\Sigma(A_0)\},$$ so $\omega(a)=1$ for every nonzero character. But then $\omega(a^*)=\overline{\omega(a)}=1$, so $\omega(a^*-a)=0$ for all characters. Similarly, $\omega(a^2-a)=0$ for all characters. Then $a^*=a=a^2$.
For part b), you have $0=a^3-a^2=0$. Then every $\lambda\in\sigma(a)$ satisfies $0=\lambda^3-\lambda^2=\lambda^2(\lambda-1)$. It follows that $\sigma(a)\subset\{0,1\}$ and so $a$ is projection by part a).