RSA: What happens to restrictions of plaintext n, dependent on p and q?
RSA has quite a few aspects, which are utilized implicitly, and these questions aim at those:


Concerning your first point about what happens if $x$ is not coprime to $n$, it does not compromise the correctness of the encryption and decryption, but if you find such an $x$, you also found a nontrivial factor of $n$. However, if $p$ and $q$ are prime, the probability to find such an $x$ is negligible (in the mathematical sense).
Concerning your first question, about what happens if $p$ and $q$ are not prime, is that you effectively weaken the scheme. For every known factor $f$ of $n$ you can reduce the RSA-problem into solving it in $f$ and $n/f$ and afterwards use the chinese remainder theorem to put them back together. An other minor aspect is, that finding $e$ might be more difficult (it has to be coprime to $\phi(n)$, otherwise decryption does not work any more, because it's not a bijection).
For your second question: If $p=q$, then $\phi(n)=\phi(p^2)=p(p-1)$ is not secret any more. Everyone can just compute $d$ by solving $ed=1$ mod $\phi(n)$. Effectively there is no trapdoor left anymore, which is required for a public key scheme to work.