Since sqrt(15)/sqrt(3)=sqrt(5), Q(sqrt(3), sqrt(5)) over Q(sqrt(15)) is the same with Q(sqrt(3), sqrt(15)) over Q(sqrt(15)). Now the basis you have is {1, sqrt(3)}.

Hint: Q(sqrt(2), cubert(2), 4thrt(2)) over Q is the same with Q(cubert(2), 4thrt(2)) over Q.[Q(sqrt(2), cubert(2), 4thrt(2)) : Q]

Any help would be appreciated. This is confusing the crap out of me.