If a vector field is divergenceless and curless, is that vector field = 0?

No. To get a feel for what's up, let $\phi$ be a non-constant harmonic function so $\nabla^2 \phi = 0$ but $\nabla \phi \ne 0$. Set $\mathbf F = \nabla \phi \ne 0$; then Then $\nabla \times \mathbf F = \nabla \times \nabla \phi = 0, \,$ since the curl of a gradient always vanishes. Also, $\nabla \cdot \mathbf F = \nabla \cdot \nabla \phi = \nabla^2 \phi = 0$. The divergence and curl of $\mathbf F$ both vanish, but not $\mathbf F$!

This line of reasoning can, like tape or film, be re-wound and run "backwards": if $\mathbf F \ne 0$ and $\nabla \times \mathbf F = 0$, then (locally at least) there is a function $\phi$ with $\mathbf F = \nabla \phi \ne 0$; if now we also have $\nabla \cdot \mathbf F = 0$, then $\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \mathbf F = 0$, and $\phi$ is harmonic.

The classic examples of such a field may be found in the elementary theory of electromagnetism: in the absence of sources, that is, charges and currents, static (non -time varying) electric fields $\mathbf E$ and magnetic fields $\mathbf B$ have vanishing divergence and curl: $\nabla \times \mathbf B = \nabla \times \mathbf E = 0$, and $\nabla \cdot \mathbf B = \nabla \cdot \mathbf E = 0$; the electrostatic potential function $\phi$ such that $\mathbf E = -\nabla \phi$ with $\nabla^2 \phi = 0$ exists by virtue of these facts; a similar assertion holds for the $\mathbf B$ field.

Hope this helps. Cheerio,

and as always (as ol' James Clerk M. has taught us),

Fiat Lux!!!

No. Just consider a nonzero constant vector field, i.e. $$V=(a,b,c)$$ for some nonzero constants $a, b, c$. Then the curl of $V$ is zero vector, and the divergence of $V$ is $0$.