= Solution
Put $D=\partial_\rho A^\rho$. The gauge-fixing variation is
$$
\delta\!\left(-\frac12D^2\right)=-D\partial^\nu\delta A_\nu.
$$
After integration by parts its contribution to the <Euler-Lagrange field equations> is $+\partial^\nu D$. It cancels the $-\partial^\nu D$ from the <Maxwell Lagrangian>, giving \b[$\Box A^\nu=0$ for all four components]. This is the <Feynman gauge> equation; the physical <Lorenz gauge> condition must still be imposed on states rather than inferred as an additional classical equation from this gauge-fixed density.
Treating the covariant components $A_\mu$ as coordinates, the derivative of the actual displayed density with respect to $\partial_0A_\mu$ gives the <canonical momenta>
$$
\boxed{\pi^\mu=-F^{0\mu}-\eta^{0\mu}D.}
$$
In terms of lower-index spatial coordinates,
$$
\pi^0=-\dot A_0+\partial_iA_i,\qquad
\pi^i=\dot A_i-\partial_iA_0\quad(i=1,2,3).
$$
The <gauge fixing> has supplied the nonzero momentum of $A_0$, absent from the ungauge-fixed <Maxwell Lagrangian>.
There is a boundary-term convention needed for the next part's printed momentum expansion. Direct expansion gives the <Feynman-gauge Maxwell kinetic density after a boundary-term subtraction>:
$$
\mathcal L=\mathcal L'+\partial_\rho K^\rho,
\qquad\mathcal L'=-\frac12\partial_\rho A_\sigma\partial^\rho A^\sigma,
\qquad K^\rho=\frac12(A_\sigma\partial^\sigma A^\rho-A^\rho D).
$$
To verify it, the difference of the densities is $\tfrac12[(\partial_\rho A_\sigma)(\partial^\sigma A^\rho)-D^2]$; differentiating $K^\rho$ gives these two terms, while the remaining second-derivative terms cancel after renaming indices. The subtracted density has $\widetilde\pi^\mu=-\dot A^\mu$, which is exactly the momentum used by the mode expansion in part (c). The literal momenta asked for here are the boxed expression, not $-\dot A^\mu$. Both densities have the same field equations, and their <canonical commutation relations> agree through the <boundary-induced canonical transformation in Feynman gauge>.
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