Two fair dice are thrown. Let X be the event that the product of the numbers is even, Y the event that the sum is a multiple of 3, Z the event that…

Over the $36$ outcomes: $P(X)=\frac34$ (product odd only when both are odd, $9$ outcomes), $P(Y)=\frac{12}{36}=\frac13$, $P(Z)=\frac{10}{36}=\frac5{18}$, $P(W)=\frac16$. Then $X\cap Y$: sums $3,6,9,12$ with an even product; of the $12$ outcomes with sum divisible by $3$ exactly $9$ have an even product, giving $P(X\cap Y)=\frac14=\frac34\cdot\frac13$, so A is true. $X\cap Z$: consecutive numbers always contain an even one, so $P=\frac5{18}\ne\frac34\cdot\frac5{18}$; B is false. $Y\cap W=\varnothing$ since $7$ is not a multiple of $3$, but $P(Y)P(W)\ne0$, so C is false. $X\cap W$: each of the $6$ outcomes with sum $7$ (one odd, one even number) has an even product, so $P=\frac16\ne\frac18$; D is false.