Is there a nice way to derive
$$ \nabla (\vec{F} \cdot \vec{G}) = \vec{F} \times (\nabla \times \vec{G} ) + \vec{G} \times (\nabla \times \vec{F}) + (\vec{F} \cdot \nabla ) \vec{G} + (\vec{G} \cdot \nabla ) \vec{F} $$
&
$$ \nabla \times (\vec{F} \times \vec{G}) = ( \vec{G} \cdot \nabla )\vec{F} - (\vec{F} \cdot \nabla ) \vec{G} + \vec{F}( \nabla \cdot \vec{G} ) - \vec{G} ( \nabla \cdot \vec{F}) $$
using differential forms?
I think I have a way to do it using differential forms operations.
First a quick note on notation, I'll use $F$ to denote both the differential one form $F_a$ as well as the vector $F^a = g^{ab} F_b$. Hopefully there will be no confusion; mostly I'll use $F$ and $G$ as vectors when doing Lie derivatives, $£_F$, or taking inner products $i_F$.
The tricky part was to come up with a way to deal with $(F\cdot\nabla)G$ and similar terms. For the first identity, we can begin by noticing that $(F\cdot\nabla)G = F^a\nabla_a G_b$ looks like the first term in the expression for the Lie derivative of a 1-form: $$£_F G = F^a\nabla_a G_b + G_a\nabla_bF^a.$$ The other term looks like one of the terms you get when you take the gradient of $F^a G_a$, i.e. $$\nabla_b(F^a G_a)=G_a\nabla_b F^a + F_a\nabla_b G^a.$$ So it is clear if we switch the roles of $F$ and $G$, we will get the other part of this term, so we arrive at the necessary identity, $$£_F G + £_G F -d(i_F G) = (F\cdot\nabla)G+(G\cdot\nabla)F. $$ Also note the following identities which are fairly straightforward to prove: $$dG = *(\nabla\times G)$$ $$i_F*H = *(H\wedge F)=-F\times H,$$ from which it follows $$-i_FdG = *(F\wedge * dG)= F\times(\nabla\times G) $$ Using these, we can prove the first identity, invoking Cartan's magic formula, $£_F = i_Fd+di_F $: $$d(i_F G) = -i_FdG+£_FG $$ $$d(i_G F) = -i_GdF+£_GF $$ Since $i_F G = i_G F = F\cdot G$, we can add these expressions to get $$ 2d(i_F G) = -i_FdG-i_GdF + £_FG+£_GF$$ $$\boxed{d(i_FG) = *(F\wedge*dG)+*(G\wedge*dF)+£_FG+£_GF-d(i_FG)} $$ This is the differential forms version of your first identity (the LHS is the gradient of the scalar $i_FG = F\cdot G$). Note that the appearance of $d(i_F G)$ on the RHS means that this expression has a more natural representation by differential forms by moving that term back over to the left, i.e. $$ d(i_F G) = \frac12(*(F\wedge*dG)+*(G\wedge*dF) + £_FG+£_GF)$$
The second identity is actually a bit cleaner. We begin by noting that for a vector $F^a$, we can represent the dual of the associated one form $F_b$ by $$*F = i_F \epsilon,$$ where $\epsilon$ is the volume $n$-form ($n=3$ for us). Using our previous identities, we can determine that the LHS is $*d*(F\wedge G)$. So we have \begin{align} *d*(F\wedge G)& = *di_G*F \\ &= *(£_G*F-i_Gd*F)\\ &=*(£_Gi_F\epsilon - i_Gd*F)\\ &=*(i_{[G,F]}\epsilon+i_F£_G\epsilon - i_Gd*F)\\ &=*(i_{[G,F]}\epsilon+i_Fdi_G\epsilon+i_Fi_Gd\epsilon-i_Gd*F) \end{align} $$\boxed{*d*(F\wedge G) = [G,F]^\flat +(*d*G)\wedge F - (*d*F)\wedge G }$$
I put a $\flat$ on the commutator of the vectors $[G,F]$ (which of course is equal to the first two terms) to remind you that it is the one form obtained by lowering the index with the metric. It is easy to verify that the $0$-form $*d*G$ is $\nabla\cdot G$. To get the fourth line I used that the Lie derivative acts as a derivation on inner products, and the final line used the fact that the exterior derivative of an $n$-form (i.e. $d\epsilon$) must vanish.
The key technique is the property of Levi-Civita symbol $\varepsilon_{ijk}$. Here list a series of formulae which is of great help. $$\varepsilon_{ijk}=\varepsilon_{jki}=\varepsilon_{kij}$$ $$\varepsilon_{ijk}\varepsilon^{ilm}=\delta^l_j\delta^m_k-\delta^l_k\delta^m_j$$ $$F\times G=\varepsilon^{ijk}F_jG_k$$
OK, now, for the first identity, $$\begin{align} \text{rhs}&=F\times(\nabla\times G)+(F\cdot\nabla)G+G\times(\nabla\times F)+(G\cdot\nabla)F\\ &=\varepsilon^{ijk}F_j\varepsilon_{klm}\partial^l G^m+F_j\partial^j G^i+\varepsilon^{ijk}G_j\varepsilon_{klm}\partial^l F^m+G_j\partial^j F^i\\ &=\varepsilon^{kij}\varepsilon_{klm}F_j\partial^l G^m+F_j\partial^j G^i+\varepsilon^{kij}\varepsilon_{klm}G_j\partial^l F^m+G_j\partial^j F^i\\ &=(\delta^i_l\delta^j_m-\delta^i_m\delta^j_l)F_j\partial^l G^m+F_j\partial^j G^i+(\delta^i_l\delta^j_m-\delta^i_m\delta^j_l)G_j\partial^l F^m+G_j\partial^j F^i\\ &=F_j\partial^iG^j+G_j\partial^iF^j\\ &=\partial^i(F_jG^j)\\ &=\nabla(F\cdot G)=\text{lhs} \end{align}$$
For the second, $$ \begin{align} \text{lhs}&=\nabla\times(F\times G)\\ &=\varepsilon^{ijk}\partial_j\varepsilon_{klm}F^l G^m\\ &=\varepsilon^{kij}\varepsilon_{klm}\partial_j(F^l G^m)\\ &=(\delta^i_l\delta^j_m-\delta^i_m\delta^j_l)\partial_j(F^l G^m)\\ &=\partial_j(F^i G^j)-\partial_j(F^j G^i)\\ &=G^j\partial_jF^i+F^i\partial_jG^j-\partial_jF^j G^i-F^j\partial_jG^i\\ &=(G\cdot\nabla)F+(\nabla\cdot G)F-(\nabla\cdot F)G-(F\cdot\nabla)G=\text{rhs} \end{align}$$
I will answer not using differential forms but using geometric calculus. It gives you a little bit more freedom, but in turn the proof does not differ from standard that much. What I see as advantage is that you don't have to do the proof in index notation.
I tried to write down answer, where I explain all the stuff needed to understand my answer(assuming knowledge of exterior calculus but not knowledge of geometric calculus) but I failed, the answer became over complicated so refer to the wiki page on geometric calculus.
Basis vectors of $\mathcal{CL^3}$ are $e_1,e_2,e_3$ with \begin{align} e_ie_j &= -e_ie_j & i\neq j \\ e_ie_j &= 1 & i=j \end{align}
Denote $F = f_i e_i, G = g_ie_i, \partial = e_i \frac{\partial}{\partial x_i} = e_i \partial_i $.
Now start differentiating $$ \nabla(F\cdot G) = \partial(F\cdot G) = \dot \partial (\dot F \cdot G) + \dot \partial(F\cdot \dot G) $$
Now we will use something similar to vector triple product in geometric algebra. That for vectors $X,Y,Z$ $$ X\cdot(Y\wedge Z) = Z(X\cdot Y) - Y(X\cdot Z) $$
No identify $X = G, Y = \dot \partial, Z = \dot F$ $$ G\cdot(\dot \partial\wedge \dot F) = \dot F(G\cdot \dot \partial) - \dot \partial(G\cdot \dot F) = (G\cdot \dot \partial) \dot F - \partial( \dot F\cdot G) $$
So we get $$ \partial( \dot F\cdot G) = (G\cdot \dot \partial) \dot F - G\cdot(\dot \partial\wedge \dot F) $$
If you do the same for $\partial( F\cdot \dot G)$ we get the final answer $$ \partial( F\cdot G) = (G\cdot \dot \partial) \dot F + (F\cdot \dot \partial) \dot G - G\cdot(\dot \partial\wedge \dot F) - F\cdot(\dot \partial\wedge \dot G) $$
If you translate those terms back to the language of vector calculus you get $$ (G\cdot \dot \partial) \dot F = (G\cdot \nabla)F $$ $$ - G\cdot(\dot \partial\wedge \dot F) = G\times ( \nabla \times F) $$
Thanks to the over-dot notation you can work with $\partial$ as with vector so you can the vector triple product formula straight away. This can not be done with standard vector calculus notation. But you can use this identity: $$ G\times (\nabla \times F) = -(G\cdot \nabla)F - G\cdot \nabla F $$ together with identity $$ \nabla (F\cdot G) = F \cdot \nabla G + G \cdot \nabla F $$ they solve your first question too.
The second identity I left you as an exercise. The $\nabla \times (F\times G)$ can be rewritten in geometric algebra as follows $$ -\partial \cdot (F\wedge G) =-\dot \partial \cdot (\dot F\wedge G) - \dot \partial \cdot ( F\wedge \dot G) $$ again you can work with $\dot \partial$ as with vector and use vector triple product formula.
