The problem is greatly simplified by defining the sequence $ C=\{c_0,c_1,c_2,\ldots\} $ 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{align*} 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{align*} 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 $ \pmod 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\pmod 4 $, we have $$c_{50} \equiv c_2 \pmod 5,$$ and thus $ c_{50}\equiv \boxed{4}\pmod 5 $.