$$\left\{\begin{aligned} a_n &&= &&2a_{n-1} + b_{n-1} + a_{n-2} - b_{n-2} && n \ge 2 && (1)\\ b_n &&=&& b_{n-1} + b_{n-2} - a_{n-2} && n \ge 2&& (2)\end{aligned}\right.$$
with $a_0 = 5, a_1 = 3, b_0 = 0, b_1 = 3$.
From (2) I get: $a_{n-2} = b_{n-1}+b_{n-2} - b_n$. Substituting in (1): $a_n = 2a_{n-1} + 2b_{n-1} - b_n$.
Now I'm stuck. I don't see what I can do next...
Hint:
Adding the two equations gives $$ a_n + b_n = 2(a_{n-1} + b_{n-1}) \quad {\rm{for}} \quad n \ge 2 $$ so $$ a_n + b_n = 2^{n-1}(a_{1} + b_{1}) = 6 \cdot 2^{n-1} $$ Plugging this into the second recursion gives $$ b_n = b_{n-1} + 2 b_{n-2} - (a_{n-2} + b_{n-2}) = b_{n-1} + 2 b_{n-2} -6 \cdot 2^{n-3} $$
Likewise,the first equation gives $$ a_n = a_{n-1} + (a_{n-1}+ b_{n-1}) + 2 a_{n-2} - (a_{n-2} + b_{n-2}) = \\ = a_{n-1} + 6 \cdot 2^{n-2} + 2 a_{n-2} - 6 \cdot 2^{n-3} = \\ = a_{n-1} + 2 a_{n-2}+ 3 \cdot 2^{n-2} $$
So there are two single-variable recursions which can be solved.