$a_{n}=a_{n-1}+b_{n-1}$ ……………………………...(1)
$b_{n}=a_{n-1}-b_{n-1}$ ……………………………………..(2)
doing $(1)+(2) $ ,
$b_{n}+a_{n}=2a_{n-1}$
So , $b_{n-1}+a_{n-1}=2a_{n-2}$
or ,$b_{n-1}=2a_{n-2}-a_{n-1}$
Substituting value of $b_{n-1}$ in (1) we get ,
$a_{n}=a_{n-1}+2a_{n-2}-a_{n-1}$
or, $a_{n}=2a_{n-2}$
Solving ,linear homogeneous equation we get ,
$r^{n}=2r^{n-2}$
or, $r^{2}-2=0$
So ,$r=\pm \sqrt{2}$
So,$a_{n}=\alpha 1\left ( \sqrt{2} \right )^{n}+\alpha 2\left ( -\sqrt{2} \right )^{n}$,
Now given , $a_{0}=1$,$a_{1}=3$,
So ,$\alpha 1$+$\alpha 2=1$ …..…..(3)
$\sqrt{2} \alpha 1-\sqrt{2}\alpha 2=3$…...……..(4)
So , $\alpha 1=\frac{\sqrt{2}+3}{2\sqrt{2}}$,$\alpha 2=\frac{\sqrt{2}-3}{2\sqrt{2}}$,
So, $a_{n}=\frac{\sqrt{2}+3}{2\sqrt{2}}(\sqrt{2})^{n}+\frac{\sqrt{2}-3}{2\sqrt{2}}(-\sqrt{2})^{n}$
$b_{n}=2a_{n}-a_{n}$ you can easily found by substituting $a_{n}$ .