(1) relation this $a_{n+1}+a_n=2^n$.With $a_0=0$.
What i did was:
multiply by 2 and it is $2a_{n+1}+2a_n=2\times 2^n$ this is (2) and on the first i did $a_{n+2} +a_{n+1}=2^{n+1}$.
I (2)-(3) and i have
$$2a_{n+1}+2a_n-a_{n+2}-a_{n+1}=0 \\2a_1+2a_0-a_{2}-a_1=0 \\2\times1+0-2-1=0 \\-1$$
Since it says to solve the recursive equation, this means you must find a formula for $a_n$, for general $n$. This is done, as you did, by getting an equation $$2a_{n+1}+2a_n-a_{n+2}-a_{n+1}=0\\\implies a_{n+2}-a_{n+1}-2a_n=0$$Now try a solution of the form $a_n=p^n$ for a constant $p$. Then $$p^2-p-2=0\\\implies (p-2)(p+1)=0\\\implies p=2,\,p=-1$$ So we get a general solution as a superposition of the two solutions, $$a_n=A 2^n+B(-1)^n$$ You can then sub in the conditions for $a_0$ and $a_1$ to get rid of the constants $A$ and $B$. Can you give that a go?