Show that regular languages are closed under Mix operations
Take any string over the alphabet $\Sigma$. Colour some of the letters red and the remaining letters green. Accept the original string if the red letters form a string in $L_1$ and the green ones are in $L_2$.

More technically, regular languages are closed onder intersection and under (inverse) homomorpisms.

Let $\Gamma = \Sigma \times \{1,2\}$ be an alphabet with colours. 

Define the homomorphisms $h_i:\Gamma^* \to\Sigma^*$, $i=1,2$ by $h_i(a,i) = a$ and $h_i(a,j) = \varepsilon$ for $i\neq j$. These keep only letters of their assigned colour, whereas their inverses colour the letters into that asigned colour while inserting arbitrary letters of the other colour.

Finally let $g:\Gamma^* \to\Sigma^*$ be the morphism $g(a,i) = a$ (which removes the colour).

Then $Mix_2(L_1,L_2) = g(\; h_1^{-1}(L_1) \cap h_2^{-1}(L_2) \;)$.

PS. Let $R_{12}$ be the regular language $[(\Sigma\times \{1\})(\Sigma\times \{2\})]^*$.
Then $Mix_1(L_1,L_2) = g(\; h_1^{-1}(L_1) \cap h_2^{-1}(L_2) \cap R_{12} \;)$.

Just for fun,  $L_1 \cdot L_2 = g(\; h_1^{-1}(L_1) \cap h_2^{-1}(L_2) \cap (\Sigma\times \{1\})^*(\Sigma\times \{2\})^* \;)$. Regular languages are closed under concatenation :)