This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and compute the nominator with the value of square root 2 and 3 memorized. And as to $e\cdot 2^{1/3}$, I just don't see how to compare it to other two items without take cubic and compute. This whole process is very time consuming.
I have seen some tricks to compare say $2^\pi,\pi^2$. But the technique does not seem to apply here.