jmc

The problem is greatly simplified by defining the sequence $C=\{c_0,c_1,c_2,\dots \}$ as $c_n=a_n+b_n$ for all nonnegative integers $n$. Then $c_0=a_0+b_0=0+1=1$ and $c_1=a_1+b_1=1+2=3$. Additionally, for integers $n>1$ we have $$\begin{array}{rl}{c}_n& =a_n+b_n\\ & =(a_{n-1}+b_{n-2})+(a_{n-2}+b_{n-1})\\ & =(a_{n-2}+b_{n-2})+(a_{n-1}+b_{n-1})\\ & =c_{n-2}+c_{n-1}.\end{array}$$ This is convenient since we want to determine the remainder of $a_{50}+b_{50}=c_{50}$. Thus, we no longer have to think about the sequences $A$ and $B$, but only about $C$.

The first few terms of $C$ are $1,3,4,7,11,18,29$. When reduced modulo $5$, these terms are $1,3,4,2,1,3,4$. The first four terms are $1,3,4,2$. These continue repeating $(\mathrm{mod}5)$ because the next two terms are $1,3$ and all terms are defined as the sum of the preceding two. Since the cycle has length $4$ and $50\equiv 2(\mathrm{mod}4)$, we have $$c_{50}\equiv c_2(\mathrm{mod}5),$$ and thus $c_{50}\equiv \boxed{4}(\mathrm{mod}5)$.