Let the predicate letters M (unary) and E (binary) respectively stand for "Is a mouse" and "Eats"
1: $\neg \exists z (Mz \wedge Ezz)$
There is no mouse that eats itself
2: $\forall x(Mx \rightarrow \forall y(My \rightarrow \neg Exy))$
Mice do not eat each other
3: $\forall x (Mx \rightarrow \forall y (Exy \rightarrow My))$
This one has me stumped! For all mice that gets eaten, it is done by other mice? If x is a mouse, if it gets eaten by something, then that other something is a mouse? If a mouse gets eaten by all things that eat, its a mouse?
Let the unary predicate letters $V,K, L$ respectively stand for "is vegetarian", "is cheese", and "is meat" and the binary predicate letter $E$ stand for "eat".
1: There is a vegetarian that doesn't eat cheese
$\exists x (Vx \wedge \forall y (Ky \rightarrow \neg Exy))$
But can I also say:
$\forall x (Kx \rightarrow \exists y (Vy \wedge \neg Eyx))$?
2: Who doesn't eat meat, is a vegetarian
Again, this one has me stumped. I would write it like this:
$\forall x \forall y ((Ly \wedge \neg E(x, y)) \rightarrow Vx)$
The first half looks good, and (3) says that mice eat only other mice. As regards the second half, I'd offer the following formulas (close to what you have, except the one I mentioned in the comments).
$\exists x(Vx ~\land \forall y(Exy \rightarrow \lnot Ky)$ ("someone is a vegetarian s.t. everything he/she eats isn't cheese").
$\forall x: Vx \lor \exists y(Ly \land Exy)$ ("everyone is either a vegetarian or there is a meat he/she eats").