Let us consider $\mathscr D(\mathbb R^3;\mathbb R^3)$ the space of smooth compactly supported vector fields in $\mathbb R^3$ and let us define
$\widetilde{\mathscr D}=\mathscr D(\mathbb R^3;\mathbb R^3)\cap \ker\text{div}$
the space of smooth compactly supported vector fields in $\mathbb R^3$ with null divergence.

Question 1: is it true that
$
\widetilde{\mathscr D}=\text{curl }
\mathscr D(\mathbb R^3;\mathbb R^3)
$?

Question 2: is it true that the topological dual ${(\widetilde{\mathscr D})}{'}$ of $\widetilde{\mathscr D}$ satisfies
$$
 {(\widetilde{\mathscr D})}{'}=\mathscr D'(\mathbb R^3;\mathbb R^3)/\text{grad }
\mathscr D'(\mathbb R^3;\mathbb R)? 
$$
These questions are implicitly important to define weak solutions of Euler or Navier-Stokes equations.
As suggested by the answers below (thanks), the first question should have a positive answer, thanks to the Poincaré lemma and the ellipticity of the exterior differentiation.

For the second question, the dual space of this subspace of $\mathscr D(\mathbb R^{3};\mathbb R^{3})$ should be the quotient
$$
\mathscr D'(\mathbb R^{3};\mathbb R^{3})/\bigl(\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})\bigr)^{\perp}.
$$
The orthogonal of $\text{curl} \mathscr D(\mathbb R^{3};\mathbb R^{3})$
is made with distributions $T$ such that, with duality brackets
$$\forall \phi\in  \mathscr D(\mathbb R^{3};\mathbb R^{3}),\quad 
0=\langle{T},{\text{curl } \phi}\rangle=\langle{\text{curl }T},{\phi}\rangle,
$$
so it is the kernel of the curl , that is
$\text{grad } \mathscr D'(\mathbb R^{3};\mathbb R)$.
Thus we should be able to infer that
$$
\bigl(\mathscr D(\mathbb R^{3};\mathbb R^{3})\cap \ker\text{div}\bigr)^{*}=\mathscr D'(\mathbb R^{3};\mathbb R^{3}){\large/}\bigl(\text{grad }\mathscr D'(\mathbb R^{3};\mathbb R)\bigr).
$$
Interestingly, if this is true, it means that when you formulate weakly the Navier-Stokes equations and you restrict the test functions to be with null divergence, as it is done in the proof of the classical Leray theorem, you obtain a solution modulo a gradient, which may be bothering (?) A few words about the latter topic: If you test only against smooth compactly supported vector fields with null divergence, you obtain that the weak limit obtained by Leray's theorem satisfies
$$
\partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=\nabla q,
$$
where $\mathbb P$ is the Leray projector. Of course, you are saved by the fact that you know that the weak limit has also null divergence, so that applying $\mathbb P$, you get 
$\partial_tu+\mathbb P (u\cdot \nabla) u+ \nu \text{ curl}^2 u=0$. My point is that this additional piece of information (div $u=0$) is due to a (linear) limit of approximate solutions, which is somehow independent of this idea of testing only against vf with null divergence. More generally, I want to point out that some information is lost when you test only against vf with null divergence.