Proof that $\nabla\bullet(f(\nabla g\times \nabla h))=\nabla f \bullet(\nabla g \times \nabla h)$. When $f,g$ and $h$ are smooth scalarfields.
Can I expand $\nabla\bullet \overbrace{(f(\nabla g\times \nabla h))}^{\text{this part}}$ like: $$\begin{align}\nabla\bullet (f(\nabla g\times \nabla h)) &= (\nabla g\times \nabla h)\bullet\nabla f+ \underbrace{f(\nabla \bullet (\nabla g\times \nabla h))}_{\text{This goes to zero it seems}} \\ &= (\nabla g\times \nabla h)\bullet\nabla f+f\bigg(\underbrace{(\nabla\times\nabla g)}_{\textbf{curl grad =0}}\bullet \nabla h-\nabla g \bullet \underbrace{(\nabla \times \nabla h)}_{\textbf{curl grad =0}}\bigg) \\ &=\nabla f \bullet (\nabla g \times \nabla h)\end{align} $$
$$\nabla\cdot({\bf F}\times{\bf G})=(\nabla\times{\bf F})\cdot {\bf G}-{\bf F}\cdot(\nabla\times{\bf G})$$
How did I get this? A triple vector product of regular vectors can be cyclically permuted without changing the result. Here, the same applies, but you need to pick the representation, in which the derivative stands next to the correct term. A derivative of a product requires one term where F is differentiated + another term where G is differentiated. In the second case, ${\bf G}\times \nabla$ doesn't make it obvious that ${\bf G}$ is being differentiated, so I reversed it.
The first one actually follows the same logic. Generate two terms in which each part is affected by the derivative, but other than which term is differentiated, the normal vector algebra rules apply. Like this:
$$\nabla \cdot (f{\bf A})=(\nabla f)\cdot{\bf A}+f(\nabla \cdot {\bf A})$$
Now that you have this, the second part can be further expanded using the rule that we used for (2) because your ${\bf A}$ is actually a cross product of two gradients.
All this is even easier, if you adopt the index notation for the vectors. If you do that, you can focus on what differentiates what, because the order of terms is no longer important for the vector notation - the indices take care of that. In this notation (Einstein summation convention is used), the problem (2) becomes
$$\epsilon_{ijk}\nabla_i (F_j G_k)=\epsilon_{ijk}(\nabla_i F_j) G_k+\epsilon_{ijk} F_j (\nabla_iG_k)$$
where $\epsilon_{ijk}$ is the Levi-Civita totally antisymmetric tensor.