= Solution
Denote the momentum in the printed mode expansion by $\widetilde\pi^\nu$, as it equals $-\dot A^\nu$ for the boundary-subtracted density discussed in part (b). Take a real complete basis of <polarization vectors>. Its completeness relation with one index raised is
$$
\sum_{\lambda,\lambda'}\epsilon_\mu^\lambda(\mathbf p)
\epsilon^{\nu\lambda'}(\mathbf p)\eta^{\lambda\lambda'}=\delta_\mu{}^\nu.
$$
Here raising or lowering the polarization labels uses the same diagonal matrix $\eta$, so this is just the completeness relation in the PDF. Write $\mathbf r=\mathbf x-\mathbf y$. In the mixed field-momentum <commutator>, the $aa^\dagger$ term gives the product of the minus sign in $\widetilde\pi$'s creation part and the minus sign in the oscillator <commutator>. The $a^\dagger a$ term gives the same sign. The two mode normalization factors multiply to $1/2$, and the momentum delta function removes one integral. Consequently
$$
\begin{aligned}
[A_\mu(\mathbf x),\widetilde\pi^\nu(\mathbf y)]
&=\frac i2\int\frac{d^3p}{(2\pi)^3}
\sum_{\lambda,\lambda'}\epsilon_\mu^\lambda\epsilon^{\nu\lambda'}\eta^{\lambda\lambda'}
\left(e^{i\mathbf p\cdot\mathbf r}+e^{-i\mathbf p\cdot\mathbf r}\right)\\
&=i\delta_\mu{}^\nu\delta^{(3)}(\mathbf x-\mathbf y).
\end{aligned}
$$
For the field-field <commutator>, the polarization sum gives instead
$$
[A_\mu(\mathbf x),A_\nu(\mathbf y)]
=-\eta_{\mu\nu}\int\frac{d^3p}{(2\pi)^3}\frac{e^{i\mathbf p\cdot\mathbf r}-e^{-i\mathbf p\cdot\mathbf r}}{2|\mathbf p|}=0,
$$
because the integrand is odd under $\mathbf p\mapsto-\mathbf p$. Similarly,
$$
[\widetilde\pi^\mu(\mathbf x),\widetilde\pi^\nu(\mathbf y)]
=-\eta^{\mu\nu}\int\frac{d^3p}{(2\pi)^3}\frac{|\mathbf p|}{2}
\left(e^{i\mathbf p\cdot\mathbf r}-e^{-i\mathbf p\cdot\mathbf r}\right)=0.
$$
These are the requested <canonical commutation relations>, derived by <photon oscillator completeness and canonical brackets>.
They also hold for the literal momenta of part (b). The shifts are
$$
\pi^0=\widetilde\pi^0+\partial_iA_i,\qquad
\pi^i=\widetilde\pi^i-\partial_iA_0.
$$
Since the fields commute at equal times, their mixed brackets with the momenta do not change. The only potentially new momentum bracket is
$$
[\pi^0(\mathbf x),\pi^i(\mathbf y)]
=i\bigl(\partial_{y^i}+\partial_{x^i}\bigr)\delta^{(3)}(\mathbf x-\mathbf y)=0.
$$
All other momentum brackets vanish separately. Thus \b[both canonical conventions have $[A_\mu,\pi^\nu]=i\delta_\mu{}^\nu\delta^3$ and vanishing coordinate-coordinate and momentum-momentum brackets], even though the momentum operators differ by spatial derivatives. This resolves the density/expansion mismatch without changing the printed oscillator algebra.
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