Solution
= Solution
Using
$$
\partial_\mu F^{\mu\nu}
=\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)
$$
and writing
$$
\alpha=\frac1{c^2}\frac{\partial\phi}{\partial t}+\nabla\cdot A,
$$
the temporal and spatial components give, with the stated metric conventions,
$$
\boxed{
\Box\phi+\frac{\partial\alpha}{\partial t}-m^2\phi
=-c\mu_0J^0,}
$$
$$
\boxed{
\Box A-\nabla\alpha-m^2A=-\mu_0J.}
$$
For $m>0$, part (c) sets $\alpha=0$. For $m=0$, $\alpha=0$ is the Lorenz gauge condition.