= Solution
If $Hv=\lambda v$ for a <Hermitian matrix> and a nonzero <eigenvector>, then $v^\dagger Hv=\lambda v^\dagger v$. The <scalar> on the left equals its complex conjugate because $H^\dagger=H$, while $v^\dagger v>0$. Thus $\lambda$ is real. If also $Hw=\mu w$, then
$$
\mu v^\dagger w=v^\dagger Hw=(Hv)^\dagger w=\lambda v^\dagger w.
$$
Consequently distinct <eigenvalues> force $v^\dagger w=0$, proving <orthogonality> of their <eigenvectors>.
For a <unitary matrix>, $U^\dagger U=I$ preserves the <norm> induced by the standard <Hermitian inner product>. If $Uv=\zeta v$ with $v\ne0$, then $\|v\|^2=\|Uv\|^2=|\zeta|^2\|v\|^2$, so $\boxed{|\zeta|=1}$.
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