jmc

(This solution is similar to Solution 2.) Let's make this a problem with boxes. In total, there are 9 products derived from these numbers (because 3 numbers per box). This would be our denominator. Every time a spinner lands on 2, we get an even product. These are $(1,2)$, $(2,1)$, $(2,2)$, $(2,3)$, and finally $(3,2)$. Going back, we see there are $3\cdot 3=9$ possible combinations, and we have $5$ evens, the final answer is $\boxed{\frac{5}{9}}$.