= Solution
Use the <Levi-Civita symbol> and the product rule. First,
$$
\begin{aligned}
\nabla\cdot(F\times G)
+&=\partial_i(\epsilon_{ijk}F_jG_k)\\
+&=\epsilon_{ijk}(\partial_iF_j)G_k
++\epsilon_{ijk}F_j\partial_iG_k\\
+&=G\cdot(\nabla\times F)-F\cdot(\nabla\times G).
\end{aligned}
$$
For the second identity, the contraction
$$
\epsilon_{ijk}\epsilon_{klm}
+=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}
$$
gives
$$
\begin{aligned}
+[\nabla\times(F\times G)]_i
+&=\epsilon_{ijk}\partial_j(\epsilon_{klm}F_lG_m)\\
+&=\partial_j(F_iG_j-F_jG_i)\\
+&=F_i\partial_jG_j-G_i\partial_jF_j
++G_j\partial_jF_i-F_j\partial_jG_i.
\end{aligned}
$$
In <vector> notation this is
$$
\boxed{
+\nabla\times(F\times G)
+=F(\nabla\cdot G)-G(\nabla\cdot F)
++(G\cdot\nabla)F-(F\cdot\nabla)G}.
$$
These are the <divergence and curl of a cross product> identities.
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