= Solution
Use the standard boundaryless-manifold convention, so $M$ is a <closed manifold>. Its orientation and <Riemannian metric> determine a unique positive <Riemannian volume form> $\mu_g$ taking value one on every positively oriented <orthonormal frame>. In an <oriented atlas> it is
$$
\mu_g=\sqrt{\det(g_{ij})}\,dx^1\wedge\cdots\wedge dx^n.
$$
For the <coordinate invariance of the Riemannian volume form>, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^Tg_yJ$ and $dy^1\wedge\cdots\wedge dy^n=(\det J)dx^1\wedge\cdots\wedge dx^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define $\mu_g$ globally.
For real $r$-forms the <Hodge star operator> is defined by
$$
\alpha\wedge*\beta=\langle\alpha,\beta\rangle_g\mu_g.
$$
The pointwise wedge pairing is <nondegenerate>, so this uniquely defines a smooth map $*:\Omega^r(M)\to\Omega^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives
$$
*^2\alpha=(-1)^{r(n-r)}\alpha.
$$
For <Hodge star eigenvalues on graded differential forms>, extend $*$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an <endomorphism> only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so $*^2=I$ and only $\pm1$ can occur. Both occur, with eigenforms $1+\mu_g$ and $1-\mu_g$.
In positive even dimension, $*^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence $*^4=I$ and the only possible <eigenvalues> are $\pm1,\pm i$. The first two occur as above. For the latter, choose a smooth local <orthonormal frame> and multiply its first dual <differential one-form> by a nonzero <smooth bump function> supported in the chart. This gives a form $\eta$ for which $\eta$ and $*\eta$ are independent. For $\lambda=\pm i$,
$$
*\bigl(\eta+\lambda^{-1}*\eta\bigr)
=\lambda\bigl(\eta+\lambda^{-1}*\eta\bigr).
$$
Thus all four occur. Consequently, for $n\geq1$,
$$
\boxed{\operatorname{Spec}(*)=
\begin{cases}
\{1,-1\},&n\text{ odd},\\
\{1,-1,i,-i\},&n\text{ even}.
\end{cases}}
$$
On degree $m$ in dimension $2m$, the <Hodge star> has <eigenvalues> $\pm1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\geq1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.
For <Hodge integration by parts in arbitrary degree>, let $\alpha\in\Omega^{r-1}(M)$ and $\beta\in\Omega^r(M)$. The <graded Leibniz rule> and <Stokes theorem> on the closed manifold give
$$
0=\int_M d(\alpha\wedge*\beta)
=\int_Md\alpha\wedge*\beta+(-1)^{r-1}\int_M\alpha\wedge d*\beta.
$$
Using the given <codifferential> and the square of the <Hodge star>,
$$
*\delta\beta
=(-1)^{n(r+1)+1+(n-r+1)(r-1)}d*\beta
=(-1)^r d*\beta.
$$
Therefore
$$
\boxed{\langle d\alpha,\beta\rangle_{L^2}
=\int_Md\alpha\wedge*\beta
=\int_M\alpha\wedge*\delta\beta
=\langle\alpha,\delta\beta\rangle_{L^2}},
$$
which proves that $\delta$ is the <formal adjoint> of $d$. The complex version inserts conjugation in the usual <Hermitian inner product>.
The <Hodge decomposition theorem> gives an $L^2$-orthogonal decomposition
$$
\Omega^k(M)=\mathcal H^k(M)\oplus d\Omega^{k-1}(M)\oplus\delta\Omega^{k+1}(M),
$$
where $\mathcal H^k(M)=\ker\Delta$ is finite dimensional and consists of the <harmonic differential forms>. Each <de Rham cohomology> class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal H}$ there is a smooth <Green operator of the Hodge Laplacian> $G$ satisfying $\Delta G=G\Delta=I-P_{\mathcal H}$.
The <solvability condition for the Hodge Poisson equation> is
$$
\boxed{\Delta\alpha=\beta\text{ is solvable}\quad\Longleftrightarrow\quad
\langle\beta,h\rangle_{L^2}=0\text{ for every }h\in\mathcal H^k(M)}.
$$
Necessity follows from $\langle\Delta\alpha,h\rangle=\langle\alpha,\Delta h\rangle=0$. Conversely the orthogonality condition gives $P_{\mathcal H}\beta=0$, so $\alpha_0=G\beta$ solves the <Poisson equation for differential forms>.
The <nonnegativity of the Hodge Laplacian> identity
$$
\langle\Delta\eta,\eta\rangle_{L^2}=\|d\eta\|_{L^2}^2+\|\delta\eta\|_{L^2}^2
$$
identifies its kernel with closed and coclosed forms. Thus the <affine space of solutions of the Hodge Poisson equation> is
$$
\boxed{\alpha_0+\mathcal H^k(M)}.
$$
For completeness, if a closed form is decomposed as $\omega=h+d\xi+\delta\zeta$, then $d\delta\zeta=0$, and $\|\delta\zeta\|^2=\langle\zeta,d\delta\zeta\rangle=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\|h\|^2=\langle\delta h,\xi\rangle=0$. This proves the canonical <isomorphism> $\mathcal H^k(M)\cong H^k_{\mathrm{dR}}(M)$, so the solution set is an <affine space> whose translation <vector space> is isomorphic to the requested <de Rham cohomology>.
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