Regularity of language obtained by interleaving $a,b$
There are several issues here. Let's first see what happens if you need to insert only $a$-s. We will construct an operator $A[r]$ which generates a regular expression for $$\{\sigma_1 a \sigma_2 a \ldots a \sigma_n : \sigma_1 \ldots \sigma_n \in L[r], n \geq 1\}.$$ If $\epsilon \in L[r]$ (which can be determined inductively), then it would need to be added explicitly as an additional final step. The construction is inductive:

$A[\epsilon] = A[\emptyset] = \emptyset$, $A[\sigma] = \sigma$.
$A[r_1 + r_2] = A[r_1] + A[r_2]$.
$A[r_1r_2] = A[r_1] a A[r_2] \stackrel?+ A[r_1] \stackrel?+ A[r_2]$, where the first additional term is present if $\epsilon \in L[r_2]$, and the second if $\epsilon \in L[r_1]$.
$A[r^*] = (A[r]a)^*A[r]$.

When you have both $a$-s and $b$-s the general idea is the same, but you have to keep track of the parity of the generated words. We construct operators $E[r],O[r]$ for extracting only the words of even length and odd length (respectively) in $L[r]$. The construction is inductive:

$E[\epsilon] = \epsilon$, $O[\epsilon] = \emptyset$.
$E[\sigma] = \emptyset$, $O[\sigma] = \sigma$.
$E[\emptyset] = O[\emptyset] = \emptyset$.
$E[r_1 + r_2] = E[r_1] + E[r_2]$, $O[r_1 + r_2] = O[r_1] + O[r_2]$.
$E[r_1r_2] = E[r_1]E[r_2] + O[r_1]O[r_2]$, $O[r_1r_2] = O[r_1]E[r_2] + E[r_1]O[r_2]$.
$E[r^*] = E[r]^*(O[r]E[r]^*O[r])^*E[r]^*$, $O[r^*] = E[r^*] O[r] E[r^*]$.

At this point we have all the ideas required to solve your actual exercise. That's a task for you.