= Solution
The <exterior algebra> of a vector space $V$ is $\Lambda^*V^*=\bigoplus_{r=0}^{\dim V}\Lambda^rV^*$, whose degree-$r$ elements are alternating $r$-linear forms. Its <exterior product> satisfies $\alpha\wedge\beta=(-1)^{rs}\beta\wedge\alpha$ for degrees $r,s$. For example, one-forms anticommute, so $dx^i\wedge dx^i=0$. A <differential form> of degree $r$ on a <smooth manifold> is a smooth section of $\Lambda^rT^*M$; in coordinates it is an alternating sum of products of $r$ coordinate one-forms. Its <exterior derivative> increases degree by one, satisfies $d^2=0$, and obeys
$$
d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^r\alpha\wedge d\beta.
$$
The coordinate definition and <pullback of a differential form> are compatible, making differential forms suitable for coordinate-independent integration.
An orientation and a nondegenerate <metric tensor> define the <Hodge star operator> by $\alpha\wedge*\beta=\langle\alpha,\beta\rangle\,\mathrm{vol}_g$ for equal-degree forms. It maps degree $r$ to degree $d-r$. If the metric has $s$ negative directions, its square is $*^2=(-1)^{r(d-r)+s}$. Thus Hodge duality turns antisymmetric tensors into complementary-degree tensors, using the metric and orientation rather than merely the underlying vector space. The <Generalized Stokes theorem> says $\int_Md\eta=\int_{\partial M}\eta$, with the induced boundary orientation, for appropriate smooth forms and compactness or support assumptions. It generalizes the fundamental theorem of calculus, circulation version of <Stokes theorem>, and divergence theorem.
For the field calculation, choose signature $(-,+,+)$ and a fixed orientation on <Minkowski spacetime>. Write $a=\delta A$, so $\delta F=da$. Symmetry of the metric pairing and the <Hodge star operator> gives $\delta(F\wedge*F)=2da\wedge*F$. Applying the <exterior derivative> product rule,
$$
da\wedge*F=d(a\wedge*F)+a\wedge d*F,\qquad
A\wedge da=F\wedge a-d(A\wedge a).
$$
Since $F\wedge a=a\wedge F$, variation of the action is
$$
\delta S=\int_M a\wedge(d*F+kF)
+\int_{\partial M}\left(a\wedge*F-\frac k2A\wedge a\right).
$$
Fix the boundary data or use compactly supported variations. Arbitrary interior one-forms $a$ then imply \b[the bulk field equation]
$$
\boxed{d*F+kF=0.}
$$
With $B=*F$, the Lorentzian <Hodge star operator> squares to $-1$ on both one- and two-forms, hence $F=-*B$. Thus the <Maxwell–Chern–Simons one-form equation> becomes
$$
\boxed{dB-k*B=0,\qquad *dB+kB=0.}
$$
These two displayed equations are equivalent. Also $dF=0$ gives $d*B=0$. For $k\ne0$ the latter follows from applying $d$ to the field equation; for $k=0$ it must be retained as the identity inherited from $F=dA$. Using the opposite metric-sign convention changes the corresponding $*^2$ sign; the signature-independent equation is $d*F+kF=0$.
Under the Abelian <gauge transformation> $A\mapsto A+d\Lambda$, the curvature $F=dA$ and the one-form $B$ are unchanged because $d^2\Lambda=0$. The field equation is therefore exactly <gauge-invariant>. The kinetic term is unchanged, while the other term changes by
$$
\boxed{\Delta S=\frac k2\int_Md\Lambda\wedge F
=\frac k2\int_Md(\Lambda F)
=\frac k2\int_{\partial M}\Lambda F.}
$$
The last equality uses $dF=0$ and the <Generalized Stokes theorem>. Thus the action density changes by a total derivative and the integrated action can change on a boundary. The source's statement that the action is not invariant needs this qualification: \b[on a boundaryless spacetime, or when the boundary integral vanishes, the integrated action is invariant too]. No such boundary restriction is needed for invariance of the bulk field equation.
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