How can I prove $\nabla \cdot \color{green}{(\mathbf{F} {\times} \mathbf{G})} $ with Einstein Summation Notation?
You are contracting the indices of $\mathbf F$ and $\mathbf G$ with each other; In $\mathbf G\cdot(\nabla\times\mathbf F)$ you'd contract it with the remaining index of $\epsilon$ (the indices of $\mathbf F$ and $\nabla$ are already contracted). However $\mathbf G\times\nabla \mathbf F$ does not make sense to me. $\partial_jF_i$ are the components of a rank-2 tensor; it might make sense to write that tensor as $\nabla\mathbf F$. However what doesn't make sense to me is to take the cross product. Looking closer to the expression, I notice that it doesn't make sense in index notation either because there are three occurrences of the index $i$ in the product; only one (free index) or two (summed index) are possible.
I don't see a difference to your first question. Just the roles of $\mathbf F$ and $\mathbf G$ are exchanged.
Your very first step is already wrong. You can easily see that by the fact that on the left hand side, $j$ is a free index, while on the right hand side, it is summed over (it appears twice in a product). A second mistake is the index $h$ which doesn't even appear on the LHS. A third mistake is the three occurrences of the index $i$.
Actually, I now even notice that your "zeroth" step also is wrong: The expression you are taking the $j$-th component from is a scalar, and therefore doesn't have a $j$-th component. (And yes, the index $j$ does mean the j-th component; anyway, considering the $j$-th term in isolation wouldn't make too much sense anyway).
Edit for 3.1 and 3.2:
Unless there's a derivative, which would determine the order. This is explained in the paragraph before "edit for 3.3 and 3.4". But otherwise, No, it is not required that $\text{[the order of the indices in the Levi-Civita symbol]}$
$= \text{[the order of the components in the component expression].}$
After all, they are just scalar factors (unless there's a derivative).
It is also not strictly required that [the order of indices in the Levi-Civita Symbol] maps the order in the vector expression. However it helps to do that because it's less error prone. What definitely matters is whether such a permutation is even or odd; this corresponds to the cross product being antisymmetric.
Putting the indices in the "right" order before conversion definitely helps with making no errors.
The second term in your 3rd equation is $\epsilon_{hij}F_i\partial_hG_j$. Since you have a derivative here, you have the additional constraint that the derivative has to come in front of what it derives, which in an expression of the form $\mathbf a\cdot(\mathbf b\times\mathbf c)$ can only happen if $\mathbf b$ is the derivative, and $\mathbf c$ is what is derived.
Therefore you want the index of $\partial$ second, and the index of $G$ third, so that those two appear in the vector product in exactly that way. That is, you want $\epsilon_{ihj}$ which you indeed get by exchanging $h$ and $i$, and therefore you get the sign change.
Edit for 3.3 and 3.4:
The order of the indices in the epsilon tensor matters, because reordering may change the sign. However, for permutations without a sign change (ie even ones), this order of the indices can change without affecting the final answer. Moreover, since the cross product is NOT commutative but the dot product is, thus in the vector expression, only the order of the vectors in the cross product matters, not the order in the dot product.
The order of the components in component form doesn't matter (except for derivative operators, of course). This means you can exchange the $A_i$, $B_j$ and $C_k$ any way you like (making sure you keep the correct index on each, of course) without changing the value: $$\epsilon_{ijk}A_iB_jC_k = \epsilon_{ijk}B_jA_iC_k = \epsilon_{ijk}B_jC_kA_i = \dots$$
Now if the order fits everywhere, then there are two possibilities for interpreting $\epsilon_{ijk}A_iB_jC_k$:
$$\begin{align} \epsilon_{ijk}A_iB_jC_k & = \color{#C33602}{\epsilon_{ijk}A_i(B_jC_k)} \qquad \text{ or } \color{#766303}{\qquad \epsilon_{ijk}(A_iB_j)C_k} \\ & = \color{#C33602}{\mathbf A\cdot(\mathbf B\times\mathbf C)} \qquad \text{ or } \color{#766303}{\qquad (\mathbf A\times\mathbf B) \cdot \mathbf C = \mathbf C \cdot (\mathbf A\times\mathbf B) } \end{align}$$ These two expressions are actually equal, due to the vector identity: $$\color{#C33602}{\mathbf A\cdot(\mathbf B\times\mathbf C)}=\mathbf B\cdot(\mathbf C\times\mathbf A) = \color{#766303}{\mathbf C\cdot(\mathbf A\times\mathbf B)}$$
So on both sides you can do odd permutations which require a sign change while even permutations which do NOT. On the LHS in terms of the components, a sign change equates with an odd permutation of the indices. On the RHS in terms of the vector expression, a sign change equates with an exchange of the two factors of the cross product (preceded and/or followed by one of the equivalent changes mentioned above, of course).
However, if
♦ the order of indices matches in the Levi-Civita symbol and the corresponding vector components
♦ and the order of the vector names matches in the component expression and the vector expression,
then you are on the safe side without putting much thought into it.
By definition
$\nabla \cdot (F\times G)=\epsilon_{jik}\partial_j(F_iG_k)= \epsilon_{jik}\partial_jF_i G_k+\epsilon_{jik}F_i\partial_j G_k=(\nabla \times F) \cdot G -F\cdot (\nabla \times G),$
as $(\nabla \times F)_k=\epsilon_{kji}\partial_j F_i=\epsilon_{jik}\partial_j F_i,$ while $(\nabla \times G)_i=\epsilon_{ijk}\partial_j G_k=-\epsilon_{jik}\partial_j G_k.$
I used
$$\epsilon_{jik}=\epsilon_{kji}, $$ $$\epsilon_{ijk}=-\epsilon_{jik}. $$