= Solution
Under the combined local transformation,
$$
D'_\mu\phi'=(\partial_\mu+iqA_\mu+iq\partial_\mu\alpha)
(e^{-iq\alpha}\phi)=e^{-iq\alpha}D_\mu\phi.
$$
The <derivative> of the phase cancels the added gauge potential. The conjugate <derivative> gains the opposite phase, so their product and $|\phi|^2$ are invariant. Also
$$
F'_{\mu\nu}=F_{\mu\nu}+\partial_\mu\partial_\nu\alpha
-\partial_\nu\partial_\mu\alpha=F_{\mu\nu}.
$$
\b[Every term of the interacting <Lagrangian> is therefore invariant.]
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