= Solution
On $p$-forms in dimension $n$, the defining identity for the <Hodge star operator> gives
$$
*^2=(-1)^{p(n-p)}.
$$
For $n=4$ and $p=2$, therefore, $*^2=1$. For every $\alpha\in\Omega^2(N)$ define
$$
\alpha_+=\frac12(\alpha+*\alpha),
\qquad
\alpha_-=\frac12(\alpha-*\alpha).
$$
Then $*\alpha_+=\alpha_+$, $*\alpha_-=-\alpha_-$, and $\alpha=\alpha_++\alpha_-$. The two eigenspaces of the involution $*$ have zero intersection, which proves uniqueness. They are respectively the spaces of <self-dual differential form>[self-dual] and <anti-self-dual differential form>[anti-self-dual] two-forms.
Now suppose $N$ is compact and let $\beta$ be an <exact differential form>[exact] three-form, say $\beta=d\theta$. Apply the <Hodge decomposition theorem> to the two-form $\theta$:
$$
\theta=h+d\varphi+\delta\psi.
$$
Set $a=\delta\psi$. Then $da=d\theta=\beta$ and $\delta a=\delta^2\psi=0$. For a two-form in dimension four, $\delta=-*d*$, so $d*a=0$. The self-dual form
$$
\eta=a+*a
$$
satisfies
$$
*\eta=*a+*^2a=\eta,
\qquad
d\eta=da+d*a=\beta.
$$
Thus every exact three-form is the <exterior derivative> of a self-dual two-form.
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