jmc

The product will be a multiple of 3 if and only if at least one of the two rolls is a 3 or a 6. The probability that Juan rolls 3 or 6 is $2/8=1/4$. The probability that Juan does not roll 3 or 6, but Amal does is $(3/4)(1/3)=1/4$. Thus, the probability that the product of the rolls is a multiple of 3 is $$\frac{1}{4}+\frac{1}{4}=\boxed{\frac{1}{2}}.$$