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Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.

The positive convention for the Laplace-Beltrami operator is ${\Delta }_{+}f=\delta df$, where $\delta$ is the codifferential. It is the restriction of the Hodge Laplacian to functions. In coordinates,On a compact boundaryless Riemannian manifold, or for compactly supported functions, $⟨f,{\Delta }_{+}f⟩=\Vert df{\Vert }^2\ge 0$. The opposite $\mathrm{div}\mathrm{grad}$ convention is also standard; formulas involving the sign must specify which is used.

For a function $f$ and one-form $\alpha$, the codifferential obeys $\delta (f\alpha )=f\delta \alpha -⟨df,\alpha {⟩}_g$. This follows from the graded Leibniz rule and the definition of the Hodge star operator. Applying it to $d(fh)=fdh+hdf$ gives the product formula. In particular ${\Delta }_{+}(f^2)=2f{\Delta }_{+}f-2|df{|}_g^2$. The cross-term sign reverses for the opposite Laplace-Beltrami operator convention.

On a compact oriented Riemannian manifold of dimension four without boundary, every exact three-form has a self-dual two-form primitive. Write $\beta =d\omega$ and use the Hodge decomposition theorem to split $\omega =h+d\xi +\delta \psi$. Then $\beta =d\delta \psi$. For $a=\delta \psi$, the codifferential identity $\delta =-\ast d\ast$ and the fact ${\ast }^2=1$ on two-forms give $\ast a=-d\ast \psi$, so $d\ast a=0$. Thus $\sigma =a+\ast a$ satisfies $\ast \sigma =\sigma$ and $d\sigma =\beta$. It is twice the self-dual projection of $a$, rather than the projection itself.