$\forall x\forall y((Male(x)\wedge Male(y))\rightarrow (x=y \vee y=x)$
For every $x,y$, if both are male, then $x = y$ or $y = x$.
The second part of OR is redundant here as $x = y$ is same as $y = x$. So, this basically means there cannot be no more than 1 man in the class (either 0 or 1 man is possible).
$\exists x \exists y (Male(x)\wedge Male(y) \wedge x\neq y \wedge \forall z(Male(z) \rightarrow (z=x \vee z=y)))$
There are $x$ and $y$ such that both are men, and if there exist another man $z$, then $x = z$ or $y = z$. This means there are exactly two men in the class- no less no more.
$\forall x \forall y \forall z (Male(x) \wedge Male(y) \wedge (Male(z)) \rightarrow (x=y \wedge x=z \wedge y=z)$
For every $x,y,z$, if all are men, then all are the same. I guess $z$ is redundant here and it should mean the same as in (1).
