= Solution
Proceed by induction on $\dim V$. The zero-dimensional case has the empty <orthonormal basis>. In positive dimension, the <fundamental theorem of algebra> gives a root of the <characteristic polynomial>, hence a unit <eigenvector> $e$ of $\alpha$, say with <eigenvalue> $\lambda$. Part (ii) makes it an <eigenvector> of $\alpha^*$ as well. Thus the line spanned by $e$ and its <orthogonal complement> are invariant under both operators by the introductory adjoint argument.
On $\{e\}^\perp$, the restriction of $\alpha^*$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted <linear operator> is therefore normal. By induction it has an <orthonormal basis> of <eigenvectors>. Adjoining $e$ proves \b[an orthonormal eigenbasis exists for $V$], establishing the finite-dimensional <spectral theorem> by construction.
For the final deduction, use this <orthonormal basis> $e_j$ and write $\alpha e_j=\lambda_je_j$. Define
$$
\beta e_j=|\lambda_j|e_j,\qquad \gamma e_j=\begin{cases}(\lambda_j/|\lambda_j|)e_j,&\lambda_j\ne0,\\e_j,&\lambda_j=0.\end{cases}
$$
The real nonnegative diagonal entries make $\beta$ a <Hermitian operator>, and the modulus-one diagonal entries make $\gamma$ a <unitary operator>. Both are diagonal in the same <orthonormal basis>, so they commute and $\beta\gamma e_j=\lambda_je_j$ even when $\lambda_j=0$. This proves the required factorization without assuming invertibility.
Conversely, if $\alpha=\beta\gamma$ with $\beta^*=\beta$, $\gamma^*\gamma=\gamma\gamma^*=I$, and $\beta\gamma=\gamma\beta$, then $\alpha^*=\gamma^*\beta$ and
$$
\alpha\alpha^*=\beta^2,\qquad \alpha^*\alpha=\gamma^*\beta^2\gamma=\beta^2.
$$
Thus $\alpha$ is normal. \b[Normality is equivalent to a commuting Hermitian–unitary factorization], with the Hermitian factor even selectable nonnegative. This is the <polar decomposition> with a unitary extension on the kernel.
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