jmc

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 $$a_{32}=32a_{31}=32\cdot 31a_{30}=\cdots =32!a_0=32!.$$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 $$b_{32}=34b_{31}=34\cdot 33b_{30}=\cdots =(34\cdot 33\cdots 3)b_0=\frac{34!}{2}.$$Thus, $$\frac{b_{32}}{a_{32}}=\frac{34!/2}{32!}=\frac{34\cdot 33}{2}=\boxed{561}.$$