Yes, indeed. The electrostatic potential $V$ satisfies $$\nabla^{2}V\left(\vec{r}\right)=-\frac{\rho\left(\vec{r}\right)}{\varepsilon_{0}}$$ so in a region $U\subset\mathbb{R}^{3}$ in which $\rho=0$, one also has $$\nabla^{2}V\left(\vec{r}\right)=0$$

Yes, indeed. The electrostatic potential $V$ satisfies $$\nabla^{2}V\left(\vec{r}\right)=-\frac{\rho\left(\vec{r}\right)}{\varepsilon_{0}}$$

so in a region $U\subset\mathbb{R}^{3}$ in which $\rho=0$, one also has $$\nabla^{2}V\left(\vec{r}\right)=0$$

Note, however, that it doesn't mean you can always do something with it. For example, most theorems on Laplace's equation will require $U$ to be a connected open subset of $\mathbb{R}^{3}$ (with the topology induced by the Euclidean metric). So taking for instance $U$ to be a bunch of discrete points with no charges won't help you so much.