 Solution

If $Hv=\lambda v$ for a Hermitian matrix and a nonzero eigenvector, then $v^{†}Hv=\lambda v^{†}v$. The scalar on the left equals its complex conjugate because $H^{†}=H$, while $v^{†}v>0$. Thus $\lambda$ is real. If also $Hw=\mu w$, thenConsequently distinct eigenvalues force $v^{†}w=0$, proving orthogonality of their eigenvectors.

For a unitary matrix, $U^{†}U=I$ preserves the norm induced by the standard Hermitian inner product. If $Uv=\zeta v$ with $v\ne 0$, then $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\zeta {|}^2\Vert v{\Vert }^2$, so $\boxed{|\zeta |=1}$.