In order to negate:
1) "Between every two distinct real numbers, there is a rational number",
we have to assert that "There are two distinct real numbers such that there is no rational number between them".
It may help to formalize the statements with quantifiers:
$\forall r_1, r_2 \in \mathbb R \ (r_1 < r_2 \to \exists q \in \mathbb Q \ (r_1 < q < r_2))$.
The "procedure" to get the correct negation is simply to put the negation sign in front and then "move it inside" with the equivalences:
$\lnot \forall$ is equiv to $\exists \lnot$ and $\lnot \exists$ is equiv to$\forall \lnot$.
Thus, from:
$\lnot \ [\forall r_1, r_2 \in \mathbb R \ (r_1 < r_2 \to \exists q \in \mathbb Q \ (r_1 < q < r_2))],$
we get in the first step:
$\exists r_1, r_2 \in \mathbb R \ \lnot (r_1 < r_2 \to \exists q \in \mathbb Q \ (r_1 < q < r_2)).$
The next step is to use the propositional equivalence between $\lnot (p \to q)$ and $(p \land \lnot q)$, to get:
$\exists r_1, r_2 \in \mathbb R \ (r_1 < r_2 \land \lnot \exists q \in \mathbb Q \ (r_1 < q < r_2)).$
For:
3) "Given any $x \in \mathbb R$, there exists $n \in \mathbb N$ satisfying $(n > x)$",
we have for the negated statement:
$\lnot \forall x \in \mathbb R \ \exists n \in \mathbb N \ (n > x)$.
Applying the above equivalences we get:
$\exists x \in \mathbb R \ \forall n \in \mathbb N \ (n \le x)$.