jmc

In eight of the twelve outcomes the product is even: $1\times 2$, $2\times 1$, $2\times 2$, $2\times 3$, $3\times 2$, $4\times 1$, $4\times 2$, $4\times 3$. In four of the twelve, the product is odd: $1\times 1$, $1\times 3$, $3\times 1$, $3\times 3$. So the probability that the product is even is $\frac{8}{12}$ or $\boxed{\frac{2}{3}}$.

OR

To get an odd product, the result of both spins must be odd. The probability of odd is $\frac{1}{2}$ on Spinner $A$ and $\frac{2}{3}$ on Spinner $B$. So the probability of an odd product is $\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)=\frac{1}{3}$. The probability of an even product, then, is $1-\frac{1}{3}=\boxed{\frac{2}{3}}$.