Thanks in advance.
From the way you put the question, I assume that part (a) is no problem.
Any string in $\displaystyle a_n$ is a string in $\displaystyle a_{n-1}$ with one of 1,2, or 3 added. To avoid repetition we can use only two of three. Therefore $\displaystyle a_n = 2 a_n $, because $\displaystyle a_1 = 3$ the given result easily follows.
EDIT:
Clearly any string in $\displaystyle b_n$ with the last number removed is a string in $\displaystyle a_{n-1}$. Therefore, taking any string in $\displaystyle a_{n-1}$ that is not in $\displaystyle b_{n-1}$ and adding the same number as it begins with gives a string in $\displaystyle b_n$. Thus $\displaystyle b_n=a_{n-1}-b_{n-1}$.
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