You did well. You just stopped too soon! Now you solve $a_{n+1}=4a_{n-1}$, which is easy: $a_{2n}=0,a_{2n+1}=4^na_1$.
So from the first given equation $b_2=-a_2-a_1=-a_1$ and hence From the second given equation $b_2=b_1-3a_0=b_1=1$, so $a_1=-1$.
Now the first equation gives $b_{2n}=-a_{2n-1}=4^{n-1}=2^{2n-2}$ and $b_{2n+1}=-a_{2n+1}=4^n=2^{2n}$.
Check: $-a_{2n-1}-a_{2n}=-a_{2n-1}=2^{2n-2}=b_{2n-1}$
$-a_{2n}-a_{2n+1}=-a_{2n+1}=2^{2n}=b_{2n+1}$
$b_{2n}+3a_{2n-1}=2^{2n-2}+3\cdot2^{2n-2}=2^{2n}=b_{2n+1}$
$b_{2n+1}+3a_{2n}=2^{2n}=b_{2n+2}$
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On your first question, I have no idea where this particular problem came from, but recurrence relations often appear when trying to solve combinatorial problems. Sometimes you get two different kinds of configuration mixed up, so it is convenient to denote their numbers as $a_n,b_n$ and derive relations like those in the Q.
Best Answer
Hint: $$a_{n+1} - 3a_n = 5 \cdot 3^{n+1} = 3 (a_n-3a_{n-1})$$
Can you start from here?