= Solution
A gauge transformation is
$$
\boxed{A_\mu\longmapsto A_\mu+\partial_\mu\chi}.
$$
Because <partial derivatives> commute, the added contribution to $F_{\mu\nu}$ is
$$
\partial_\mu\partial_\nu\chi-\partial_\nu\partial_\mu\chi=0,
$$
so the <electromagnetic field tensor> is gauge invariant.
Define
$$
E=-\nabla\Phi-\partial_tA,
\qquad
B=\nabla\times A.
$$
Since $\partial_0=c^{-1}\partial_t$ and $A_0=-\Phi/c$,
$$
\boxed{F_{0i}=-\frac{E_i}{c},
\qquad
F_{i0}=\frac{E_i}{c},
\qquad
F_{ij}=\epsilon_{ijk}B_k}.
$$
The identity
$$
\partial_\lambda F_{\mu\nu}
+\partial_\mu F_{\nu\lambda}
+\partial_\nu F_{\lambda\mu}=0
$$
follows directly from $F=dA$. Its spatial and mixed components are
$$
\nabla\cdot B=0,
\qquad
\nabla\times E=-\partial_tB.
$$
For the other two equations define the four-current
$$
\boxed{j^\mu=(c\rho,J)}.
$$
Then $\partial_\nu F^{0\nu}=\mu_0j^0$ gives
$$
\nabla\cdot E=\mu_0c^2\rho=\frac\rho{\epsilon_0},
$$
and $\partial_\nu F^{i\nu}=\mu_0j^i$ gives
$$
\nabla\times B-\frac1{c^2}\partial_tE=\mu_0J.
$$
This is the <Covariant Maxwell equation with the minus-plus-plus-plus metric>.
Using
$$
F^{\rho\sigma}F_{\rho\sigma}
=2\left(B^2-\frac{E^2}{c^2}\right),
$$
one finds
$$
\boxed{
T^{00}=\frac1{2\mu_0}
\left(B^2+\frac{E^2}{c^2}\right)
=\frac12\left(\frac{B^2}{\mu_0}+\epsilon_0E^2\right)}.
$$
This is the electromagnetic energy density.
For a null <vector>, the trace term in $T^{\mu\nu}k_\mu k_\nu$ vanishes. Put
$$
q^\rho=k_\mu F^{\mu\rho}.
$$
Antisymmetry gives $q\cdot k=0$. A <vector> orthogonal to a null <vector> has nonnegative Minkowski norm, so
$$
T^{\mu\nu}k_\mu k_\nu=\frac1{\mu_0}q^\rho q_\rho\geq0.
$$
Explicitly, in a frame with $k^\mu=\kappa(1,1,0,0)$,
$$
q^2=\kappa^2\left[
\left(B_3-\frac{E_2}{c}\right)^2
+\left(B_2+\frac{E_3}{c}\right)^2
\right].
$$
Hence the <null energy condition for the electromagnetic field> is strict whenever the contraction is nonzero:
$$
\boxed{T^{\mu\nu}k_\mu k_\nu>0}.
$$
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