$a_{1}=3a_{0}+2b_{0}=3+4=7$
Now,
$a_{n}-b_{n}=3a_{n-1}-a_{n-1}$
$a_{n}-2a_{n-1}=b_{n}\ \ ................(1)$
Substituted the value of $b_{n}$ in first recurrence relation
$a_{n}=3a_{n-1}+2(a_{n-1}-2_{n-2})$
$a_{n}=5a_{n-1}-4a_{n-2}$
Now, solving this linear homogeneous recurrence relation :
$r^n=5r^{n-1}-4r^{n-2}$
Divide by $r^{n-2}$ on both sides
$r^2=5r-4$
$r^2-5r+4=0$
$r=1,4$
$a_{n}=\alpha_{1}(1)^n+\alpha_{2}(4)^n$
$a_{0}=1=\alpha_{1}+\alpha_{2}$
$a_{1}=7=\alpha_{1}+4\alpha_{2}$
Solving both equations gives : $\alpha_{1}=-1, \alpha_{2}=2$
$\therefore a_{n}=-1+2(4^n)$
Using $(1)$
$b_{n}=-1+2(4)^n-2(-1+2(4)^{n-1})$
$b_{n}=-1+2(4)^n+2-4(4)^{n-1}$
$b_{n}=1+2(4)^n-4^n$
$b_{n}=1+4^n$