Call the family of those sets $T$. Let us show that $T_0 = T \cup \{ \emptyset \}$ is a topology.
First, $\emptyset \notin T$, so we have to adjoint it to $T$ "by hand". Done.
Next, $\Bbb N \in T_0$, obviously: for every $n \in \Bbb N$, the divisors of $n$ are still in $\Bbb N$.
Let $U,V \in T_0$. If any of them is $\emptyset$, then $U \cap V = \emptyset \in T_0$. If $U,V \ne \emptyset$, then let $n \in U \cap V$. Let $m$ be a divisor of $n$. Then $m \in U$ and $m \in V$ because $U,V \in T$, so $m \in U \cap V$, therefore $U \cap V \in T$.
Let $\{U_i \mid i \in I \} \subseteq T_0$. If $U_i = \emptyset \ \forall i \in I$, then $\bigcup _{i \in I} U_i = \emptyset \in T_0$. If at least one $U_i$ is non-empty, then let $n \in \bigcup _{i \in I} U_i$. There must exist a $j \in I$, then, such that $n \in U_j$. In particular, this means that $U_j \ne \emptyset$, so $U_j \in T$. Let $m$ be a divisor of $n$. Since $U_j \in T$, then $m \in U_j$, so that $m \in \bigcup _{i \in I} U_i$, which shows that $\bigcup _{i \in I} U_i \in T \subset T_0$.
Let now $U_n = \{ m \in \Bbb N \mid m \mid n \}$ be the set of the natural divisors of $n$ . Notice that each $n \in \Bbb N$ belongs to $U_n$, so $(U_n)_{n \ge 0}$ covers $\Bbb N$. Next, think of $U_m \cap U_n$. You would like a $U_d$ such that $U_d \subseteq U_m \cap U_n$. Choose $d = \gcd (m, n)$; if $k \in U_d$, then $k \mid d$, and since $d \mid m$, then $k \mid m$ so $k \in U_m$. Similarly, $k \in U_n$, so $k \in U_m \cap U_n$, therefore $U_d \subseteq U_m \cap U_n$. It follows then that $(U_n)_{n \ge 0}$ is a base for $T_0$.
Let us now look at $B = \{ 2 \}$. The only posibilities for $\mathring B$ are $\emptyset$ and $B$. If $B$ were an open set, then all the natural divisors of $2$ should belong to it; but $1 \notin B$, so $B \notin T$, leaving as the only possibility $\mathring B = \emptyset$.
The closure of $B$ is extremely interesting and weird. Remember that $\bar B$ is made of all the limits of sequences from $B$. Fortunately, since $B$ contains a single point, there is a single sequence living in $B$, namely the constant sequence $2, 2, 2, \dots$. Unlike what you expect, the limit of this sequence is not just $2$! Consider an even number $2n$. Pick any open subset containing it; this open subset will then contain all the divisors of $2n$, so in particular it will contain $2$. Any neighbourhood of $2n$ contains the terms of the sequence! This shows that every even number is a limit point for it, which in particular shows that this topology is not Hausdorff. It follows that $\bar B = \{0, 2, 4, 6, \dots\} = \{ 2n \mid n \in \Bbb N \}$.
It immediately follows that $\partial B = \bar B \setminus \mathring B = \{ 2n \mid n \in \Bbb N \}$.
Now, remember that $B'$ is the set of limits of non-constant sequences from $B$. Since the only sequence from $B$ is constant, it follows that $B' = \emptyset$.
It follows immediately that the set of isolated points is $\bar B \setminus B' = \{ 2n \mid n \in \Bbb N \}$.
The exterior of $B$ is, by definition, $\Bbb N \setminus \bar B$ (the odd numbers).
The conclusions hold almost verbatim for every subset made of a single natural number.
Let us finally study $A$. Remember that the interior of a set is the largest open set contained in it. The subsets of $A$ are $\emptyset, \{1\}, \{3\}, \{4\}, \{1,3\}, \{1,4\}, \{3,4\}, \{1,3,4\}$. Of these, the subsets containing $4$ do not also contain $2$ (which divides $4$), so they cannot belong to the topology. Of the remaining ones, the largest one is abviously $\{1,3\}$, so $\mathring A = \{1,3\}$.
The closure is easier: just remember that the closure commutes with finite unions, so
$$\overline {\{1,3,4\}} = \overline {\{1\}} \cup \overline {\{3\}} \cup \overline {\{3\}} = \{ 1 \cdot n \mid n \in \Bbb N\} \cup \{ 3 \cdot n \mid n \in \Bbb N \} \cup \{ 4 \cdot n \mid n \in \Bbb N \} = \\
\Bbb N \cup 3 \Bbb N \cup 4 \Bbb N = \Bbb N ,$$
where for each set made of a single number I have applied the result obtained previously for $B$.
It immediately follows that $\partial A = \bar A \setminus \mathring A = \Bbb N \setminus \{1,3\}$.
The derived set, too, commutes with finite unions, so $A' = \{1\}' \cup \{3\}' \cup \{4\}' = \emptyset \cup \emptyset \cup \emptyset = \emptyset$, where again I have used the result obtained for $B$.
The isolated points are $\bar A \setminus A' = \Bbb N$.
The exterior is just $\Bbb N \setminus \bar A = \emptyset$.
