= Solution
Use the <Einstein summation convention>, with <divergence> $\partial_iF_i$ and <curl> $(\nabla\times\mathbf F)_i=\epsilon_{ijk}\partial_jF_k$, where $\epsilon_{ijk}$ is the <Levi-Civita symbol>. For the first <vector field>,
$$
\partial_jF_i=2x_jx_i+r^2\delta_{ij},\qquad
\boxed{\nabla\cdot\mathbf F=5r^2,\quad \nabla\times\mathbf F=\mathbf0.}
$$
The derivative matrix is symmetric, so its contraction with the <Levi-Civita symbol> vanishes. For the second <vector field>,
$$
\partial_jG_i=a_jx_i+(\mathbf a\cdot\mathbf x)\delta_{ij},\qquad
\boxed{\nabla\cdot\mathbf G=4\mathbf a\cdot\mathbf x,\quad \nabla\times\mathbf G=\mathbf a\times\mathbf x.}
$$
For the third <vector field>, $H_i=\epsilon_{ikl}a_kx_l/r$. Its <divergence> is zero because the <Levi-Civita symbol> contracts with symmetric products. For its <curl>, either contract two <Levi-Civita symbols> or use the <vector triple product>:
$$
\nabla\times\left(\frac{\mathbf a\times\mathbf x}{r}\right)
=-\frac{\mathbf x}{r^3}\times(\mathbf a\times\mathbf x)+\frac{2\mathbf a}{r}
=\frac{\mathbf a}{r}+\frac{(\mathbf a\cdot\mathbf x)\mathbf x}{r^3}.
$$
Thus
$$
\boxed{\nabla\cdot\mathbf H=0,\qquad
\nabla\times\mathbf H=\frac{\mathbf a+(\mathbf a\cdot\widehat{\mathbf x})\widehat{\mathbf x}}r.}
$$
All formulas hold on $r>0$, the domain where the radial <unit vector> is defined.
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