$3$. $0$ is a limit point of the subset $\{1/n : n \in \mathbb{N}\}$,
Proof:
Let $U_0$ be an arbitrary open neighborhood of $0$ of the form $(-a, \epsilon) \cap (-\infty,0]\cup\{1/n : n \in \mathbb{N}\}$, with $\epsilon > 0 $, $a>0$ subject to choice. Then for $N>\frac{1}{\epsilon}$, we have $\frac{1}{N} < \epsilon$. So then:
$$
\{\frac{1}{N},0\} \subset (-a, \epsilon) \cap (-\infty,0]\cup\{1/n : n \in \mathbb{N}\} \cap \{1/n : n \in \mathbb{N}\}
$$
i.e. the intersection contains something other than $0$, namely: $\{\frac{1}{N}\}$. So $0$ is a limit point for the set: $\{1/n : n \in \mathbb{N}\}$.
$4.$ $(-2,0)$ is open because $(-2,0)$ is open in $\mathbb{R}$ and
$$
(-2,0)\cap (-\infty,0]\cup \{1/n : n \in \mathbb{N}\} = (-2,0)
$$
is open in the subspace topology by definition.