Dividing the recurrence relation for $ \{a_n\} $ by $ a_{n-1} $, we get \[\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}}.\]Then, since $ a_1/a_0 = 1 $, we have $ a_2/a_1 = 1 + (a_1/a_0) = 2 $, $ a_3/a_2 = 1 +(a_2/a_1) = 3 $, and so on. In general, $ a_n/a_{n-1} = n $ for all $ n $. Then \[a_{32} = 32a_{31} = 32 \cdot 31a_{30} = \dots = 32! a_0 = 32!.\]For $ \{b_n\} $, we also have $ b_n/b_{n-1} = 1 + (b_{n-1}/b_{n-2}) $, but here $ b_1/b_0 = 3 $. Therefore, in general, $ b_n/b_{n-1} = n+2 $ for all $ n $. Then \[b_{32} = 34b_{31} = 34\cdot 33b_{30} = \dots = (34 \cdot 33 \cdots 3)b_0 = \frac{34!}{2}.\]Thus, \[\frac{b_{32}}{a_{32}} = \frac{34!/ 2}{32!} = \frac{34 \cdot 33}{2} = \boxed{561}.\]