From either bag, you have [tex]\dfrac36[/tex] chance of drawing an even-numbered tile. Whatever you draw from the first bag is independent of what you draw from the second, which means
[tex]\mathbb P((\text{even from bag 1})\cap(\text{even from bag 2}))=\mathbb P(\text{even from bag 1})\times\mathbb P(\text{even from bag 2})=\dfrac36\times\dfrac36=\dfrac9{36}[/tex]
