How do I attack an RSA setup where e is even?
Unless they did something wrong (either accidentally, or deliberately to make it easy), there is no practical way.

It's well known that, if you're able to compute the squareroot of an arbitrary number modulo a composite, you can efficiently factor that composite.  And, solving $e=4$ is equivalent for solving the RSA problem twice with $e=2$.

Now, it's possible that they somehow made it feasible for you, such as:


Selecting a prime $N$, or one that is easy to factor
Selecting a tiny message (however, you did indicate that they didn't do that)


As for why we generally select $e$ relatively prime to $p-1$ and $q-1$, well, that's not actually to make things more difficult for the attacker.  Instead, it's so that the legitimate decryptor (that is, the guy with the private key) can uniquely determine the original message.

After all, we have $M^4 = (N-M)^4 \bmod N$ for all messages $M$; hence if the decryptor got the value $M^e \bmod N$, he would have not inherent way to determine if $M$ or $N-M$ was the original message.

However, that doesn't help you.