 Solution

If $I+A$ were a singular matrix, a nonzero vector $v$ would satisfy $(I+A)v=0$, so $Av=-v$. This would make $-1$ an eigenvalue of the anti-Hermitian matrix, contrary to part (a). Thus $\boxed{I+A\text{ is invertible}}$. The same argument shows $I-A$ is an invertible matrix, which will also be used below.