Let be the full-shuffle operation, $S(L_1,L_2)=\{w|w=w_1w_2...w_k\mbox{ such that $w_{2k+1}\in L_1$ and $w_{2k}\in L_2$, $k\in\mathbb{N}$}\}$. That is to say the language $L$ that contains the words which we can build from a word from $L_1$, followed by a word from $L_2$, followed by a word from $L_1$, etc...
How to show that context-free languages are closed by this operation ?
My attempt
Let be two grammars which variables are disjoint and whichi initial variables are $S_1$ and $S_2$. We take the grammar which have a new initial variable $S$.
$$S\rightarrow \varepsilon | S_1|S_2$$ $$S_1\rightarrow DS_1 $$ $$S_2\rightarrow DS_2 $$ $$DS_1\rightarrow DS_1S_2 $$ $$DS_2\rightarrow DS_2S_1 $$
Just like in your previous question, there is a small ambiguity in the definition since you do not specify whether $k$ should be even. Anyway, if you assume that $k$ is even, then $S(L_1,L_2) = (L_1L_2)^*$ and if there is no restriction on $k$, then $S(L_1,L_2) = (L_1L_2)^* \cup (L_1L_2)^*L_1$.
In both cases, if $L_1$ and $L_2$ are regular, then $S(L_1,L_2)$ is regular (which answers your previous question, and if $L_1$ and $L_2$ are context-free, then $S(L_1,L_2)$ is context-free, since regular [respectively context-free] languages are closed under union, product and star.