The events of rolling a 3 or a 5 are disjoint, so [tex]\mathbb P\bigg((X=3)\cup(X=5)\bigg)=\mathbb P(X=3)+\mathbb P(X=5)[/tex].
Each face has a [tex]\dfrac16[/tex] probability of occurring, so the event of interest occurs with probability [tex]\dfrac16+\dfrac16=\dfrac13[/tex]. The number of times a 3 or 5 is rolled across 288 trials has a binomial distribution, [tex]\mathcal B\left(288,\dfrac13\right)[/tex].
Recall that the expected value of a binomial distribution [tex]\mathcal B(n,p)[/tex] is [tex]np[/tex]. So in this experiment, we should expect to witness approximately [tex]\dfrac{288}3=96[/tex] instances of rolling a 3 or a 5.
