= Solution
Use $\hbar=c=1$, signature $(+---)$, and define $D_\mu=\partial_\mu+ieA_\mu$. Under a local phase transformation,
$$
\phi'=e^{-ie\alpha(x)}\phi,\qquad
A'_\mu=A_\mu+\partial_\mu\alpha,
$$
the derivative of the phase cancels the change of the potential, giving $D'_\mu\phi'=e^{-ie\alpha}D_\mu\phi$. Therefore a gauge-invariant scalar-electromagnetic <Lagrangian> density is
$$
\boxed{\mathcal L=(D_\mu\phi)^*D^\mu\phi-m^2\phi^*\phi
-\frac14F_{\mu\nu}F^{\mu\nu},\qquad
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.}
$$
The mass term is invariant, the covariant-derivative product has canceling phases, and $F'_{\mu\nu}=F_{\mu\nu}$ because mixed <partial derivatives> commute. A potential depending only on $\phi^*\phi$ may be added without changing the symmetry. The Maxwell term makes the electromagnetic field dynamical; it can be omitted when $A_\mu$ is a prescribed external background.
Expanding the covariant-derivative term gives the interaction linear in $e$ plus \b[$e^2A_\mu A^\mu|\phi|^2$]. This quadratic term is required for exact local <gauge invariance>. The later part (c) deliberately gives only the linear interaction, and must be solved as that different stated <Lagrangian> rather than silently adding the missing term.
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