"Clearly not diagonalizable" is not correct; if we know that $N^2\neq 0$, then you are correct (that would imply that the minimal polynomial of $N$ is also $x^3$, and since it is not square free then $N$ is not diagonalizable). But just from knowing that $N$ has characteristic polynomial $x^3$, we do not know whether $N$ is diagonalizable or not. It could be diagonalizable. Explicitly, $N$ is diagonalizable if and only if it is the zero matrix; prove it!
As noted, you cannot have 2 and 4 both false, since they are negations of each other and the excluded middle applies here. And 2 and 3 are logically equivalent for a $3\times 3$ matrix.
If the question explicitly states that $N^2\neq 0$, then you know that 2 is false. If the question does not explicitly state so, then you don't.