Surely you have $K_a=\{\frac{a}{n+1}|n \in \mathbb{N}\}=\{\frac{a}{2},\frac{a}{3},\frac{a}{4},\ldots\}$
For an element to be in the interior you would need an open $\epsilon$ neightborhood of the elment to be in the set which is not the case here, as you only have isolated points. The closure of the set is the smallest superset that is closed under the convergence of sequences, so you would have to add the element $0$ because you can have sequences that converge against $0$.
Now the boundary is the closure without the interior and there you have all your sets.
Let's do the integers together and then you can try the other ones on your own:
(i) $\operatorname{int}{\mathbb Z}$:
By definition the interior of a set $S$ are all the points $x$ in it such that we can find an $\varepsilon > 0$ such that $B(x,\varepsilon) \subset S$.
Let $\varepsilon > 0$ and $n \in \mathbb Z$. Our $\varepsilon$ is small but we know that if we pick a $k \in \mathbb N$ large enough then we can achieve $\frac{1}{k} < \varepsilon$. But then $n + \frac{1}{k} \in B(n, \varepsilon)$ and $n + \frac{1}{k} \notin \mathbb Z$, so $B(n,\varepsilon) \not\subset \mathbb Z$. So the interior of $\mathbb Z$ must be empty.
(ii) $\overline{\mathbb Z} \subset \mathbb R$:
We know that $\overline{\mathbb Z} = \partial \mathbb Z \cup \operatorname{int}{\mathbb Z}$ so let's find out what the boundary $\partial \mathbb Z$ is.
(iii) $\partial \mathbb Z$:
By definition, $\partial \mathbb Z$ are all points $n$ such that every $B(n, \varepsilon)$ has non-empty intersection with both, $\mathbb Z$ and $\mathbb R \setminus \mathbb Z$. Let $n$ be any point in $\mathbb Z$ and $B(n, \varepsilon)$ an epsilon ball around it. Then the intersection $\mathbb Z \cap B(n, \varepsilon)$ contains $n$ hence is non-empty. Also, in (i) we saw that every such epsilon ball has non-empty intersection with $\mathbb R \setminus \mathbb Z$. So $\partial \mathbb Z = \mathbb Z$.
Best Answer
Hint. Let's $\tau$ is the topology of your question. You can think of a more concrete way of explaining the face of this open topology.If $O\in \tau$ and $O\neq \emptyset$ then $O^c$ is finite. That is, thare is $n$ numbers $x_1<x_2<\ldots, x_{n-1}<x_n$ such that $O^c=\{x_1, \ldots, x_n\}$. Then, $$ O=(-\infty,x_1)\cup( x_1,x_2)\cup\ldots\cup( x_{n-1},x_n)\cup(x_n,\infty) $$ This means that each nonempty open $O\in\tau$ contains at least two intervals of infinite type $$ (-\infty,a) \mbox{ and } (b,+\infty) \mbox{ with } a\leq b. $$ Therefore the only open that can be contained in the set S is empty. Since, by definition, the interior of a set $S$ is the union of all open contained in $S$ then the interior of $S$ is the empty set ($S$ contains only empty set).