= Solution
Hold the metric fixed when varying the <gauge potential>, put $F=dA$, and take variations with compact support or with appropriate fixed boundary data. The <Hodge star operator> then gives
$$
\delta S_{\rm kin}=-\int d(\delta A)\wedge *F=-\int\delta A\wedge d*F-\int d(\delta A\wedge *F).
$$
For the Abelian <Chern-Simons 5-form>, the two occurrences of $F$ commute as even-degree <differential forms>. Since $dF=0$,
$$
\delta(A\wedge F\wedge F)=\delta A\wedge F\wedge F+2A\wedge d(\delta A)\wedge F
=3\delta A\wedge F\wedge F-2d(A\wedge\delta A\wedge F).
$$
The factor three comes from all three appearances of $A$, including those hidden in $F=dA$. The bulk variation is therefore
$$
\delta S=\int\delta A\wedge(-d*F+3cF\wedge F)+\text{boundary terms},
$$
and the <Abelian Chern-Simons five-form field equation> is
$$
\boxed{d*F=3cF\wedge F,\qquad dF=0.}
$$
Under an Abelian <gauge transformation> $A\mapsto A+d\lambda$, the <gauge field strength> stays fixed and the action changes by
$$
\Delta S=c\int d\lambda\wedge F\wedge F=c\int d(\lambda F\wedge F)=c\int_{\partial M}\lambda F\wedge F.
$$
Thus the local integrand is not gauge invariant, but its change is a boundary term by the <Generalized Stokes theorem>. With the stated boundary conditions the variational bulk equations are unchanged; they depend only on $F$ and the metric, so they are manifestly gauge invariant. On a spacetime with boundary, gauge transformations nonzero at the boundary require boundary conditions or boundary degrees of freedom; the bulk claim does not assert invariance of every possible boundary action.
For the metric variation, write the kinetic action in components,
$$
S_{\rm kin}=-\frac14\int d^5x\sqrt{-g}\,F_{\rho\sigma}F^{\rho\sigma}.
$$
At fixed covariant components $A_\mu$ and $F_{\mu\nu}$,
$$
\delta\sqrt{-g}=-\frac12\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu},\qquad
\delta(F_{\rho\sigma}F^{\rho\sigma})=2F_{\mu\rho}F_\nu{}^\rho\delta g^{\mu\nu}.
$$
Substituting into the given definition of the <stress-energy tensor> yields
$$
\boxed{T_{\mu\nu}=F_{\mu\rho}F_\nu{}^\rho-\frac14g_{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.}
$$
The <Chern-Simons 5-form> uses only <wedge products of differential forms> and orientation, with no metric or <Hodge star operator>. Its bulk metric derivative is zero, so \b[the Chern-Simons term contributes no bulk stress-energy tensor]. The displayed tensor is generally not traceless in five dimensions: $T^\mu{}_\mu=-F_{\rho\sigma}F^{\rho\sigma}/4$.
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