= Solution
A <Hermitian matrix> satisfies $A^\dagger=\overline A^T=A$. If $Av=\lambda v$ with $v\ne0$, then $v^\dagger Av=\lambda v^\dagger v$ is real because its conjugate equals $v^\dagger A^\dagger v=v^\dagger Av$. Since $v^\dagger v>0$, its <eigenvalue> $\lambda$ is real. If $Aw=\mu w$ with $\mu\ne\lambda$, then
$$
\mu v^\dagger w=v^\dagger Aw=(Av)^\dagger w=\lambda v^\dagger w,
$$
so $v^\dagger w=0$: distinct eigenspaces are <orthogonal> for the standard complex <inner product>.
The converse is false. For example $\begin{pmatrix}1&1\\0&2\end{pmatrix}$ has the two real <eigenvalues> one and two but is not a <Hermitian matrix>. It can be embedded as a block in a $3\times3$ example if desired.
For Hermitian $A,B$, $(AB)^\dagger=B^\dagger A^\dagger=BA$. Thus $AB$ is Hermitian exactly when $AB=BA$, equivalently when their <commutator> vanishes. Also $(AB+BA)^\dagger=AB+BA$, so a value that works for every such pair is
$$
\boxed{c=1.}
$$
This sum is their <anticommutator>.
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