= Solution
Keep the metric and hence the <Hodge star> fixed. Write $\alpha=\delta A$, so $\delta F=d\alpha$, and use the <Bianchi identity for an Abelian p-form> $dF=d^2A=0$. The bilinear pairing $U\wedge\star V$ on real four-forms is symmetric even for a fixed indefinite metric. Consequently the variation of the kinetic term is $d\alpha\wedge\star F$.
The two differentiated four-form factors in the cubic term have the same sign, since moving a four-form through another four-form or through a three-form contributes an even exponent. Thus
$$
\delta(F\wedge F\wedge A)=2d\alpha\wedge F\wedge A+\alpha\wedge F\wedge F.
$$
The <graded Leibniz rule> gives
$$
d(\alpha\wedge\star F)=d\alpha\wedge\star F-\alpha\wedge d\star F,
\qquad
d(\alpha\wedge F\wedge A)=d\alpha\wedge F\wedge A-\alpha\wedge F\wedge F.
$$
Using the <Generalized Stokes theorem>, the complete <first variation> is therefore
$$
\delta S=\int_M\alpha\wedge(d\star F+3F\wedge F)
+\int_{\partial M}(\alpha\wedge\star F+2\alpha\wedge F\wedge A).
$$
For compactly supported variations, a closed $M$, or boundary conditions removing this last term, the <Euler-Lagrange equation> and the <Bianchi identity for an Abelian p-form> are
$$
\boxed{d\star F+3F\wedge F=0,\qquad dF=0.}
$$
The coefficient three follows from the normalization actually present in the action; inserting a differently normalized supergravity coupling would change it. This is the <eleven-dimensional three-form action variation>.
Under the <gauge transformation> $A\mapsto A+d\Lambda$, the four-form $F$ is unchanged because $d^2\Lambda=0$. The kinetic density is unchanged, and the cubic density changes by
$$
F\wedge F\wedge d\Lambda=d(F\wedge F\wedge\Lambda).
$$
It is an exact eleven-form, so the action changes only by a boundary integral. The <Euler-Lagrange equations> depend only on $F$ and are invariant even when this boundary integral is nonzero. Invariance of the action itself additionally requires suitable boundary or support conditions.
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