$\begin{aligned} {[\vec{\nabla} \times(s \vec{A})]i } & =\varepsilon{i j k} \partial_j(s \vec{A})k \ & =\varepsilon{i j k} s \partial_j A_k+\varepsilon_{i j k} A_k \partial_j s \ & =s \vec{\nabla} \times \vec{A}+\vec{\nabla} s \times \vec{A} .\end{aligned}$

The curl of an electrostatic field must be zero, but otherwise there is no restriction. So the answer follows as
(i) $\vec{\nabla} \times \vec{E}(\vec{r})=\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right) \hat{e}_z+\ldots=\overrightarrow{0}$. YES, it describes an electric field.
(ii) $\vec{\nabla} \times \vec{E}(\vec{r})=(1-1) \hat{e}_z=0$. YES, it describes an electric field.
(iii) $\vec{\nabla} \times \vec{E}(\vec{r})=(-1-1) \hat{e}_z=-2 \hat{e}_z$. NO, it does not describe an electric field.

This problem can be solving using either traceless symmetric tensors or the more standard spherical harmonics. I will show the solution both ways, starting with the simplier derivation in terms of traceless symmetric tensors.
(a) We exploit the fact that the most general solution to Laplace’s equation can be written as a sum of terms of the form
$$
\left(r^{\ell} \text { or } \frac{1}{r^{\ell+1}}\right) C_{i_1 \ldots i_{\ell}}^{(\ell)} \hat{n}{i_1} \ldots \hat{n}{i_{\ell}},
$$
where $C_{i_1 \ldots i_{\ell}}^{(\ell)}$ is a traceless symmetric tensor. In this case we only need an $\ell=1$ term, since
$$
F_a(\theta, \phi) \equiv \sin \theta \cos \phi=\frac{x}{r}=\hat{x}_i \hat{n}_i
$$
For $\ell=1$ the radial function must be $r$ or $1 / r^2$. For $rR$ the term proportional to $r$ is excluded, because it does not approach zero as $r \rightarrow \infty$, so only the $1 / r^2$ option remains, and the solution is
$$
\begin{aligned}
V(\vec{r}) & =V_0\left(\frac{R}{r}\right)^2 F_a(\theta, \phi) \
& =V_0\left(\frac{R}{r}\right)^2 \hat{x}_i \hat{n}_i \text { or } V_0\left(\frac{R}{r}\right)^2 \sin \theta \cos \phi .
\end{aligned}
$$