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.
I understand in your comment above to Jonas' answer that you would like these things to be broken down into simpler terms.
Think about limit points visually. Suppose you have a point $p$ that is a limit point of a set $E$. What does this mean? In plain terms (sans quantifiers) this means no matter what ball you draw about $p$, that ball will always contain a point of $E$ different from $p.$
For example, look at Jonas' first example above. What you should do wherever you are now is draw the number line, the point $0$, and then points of the set that Jonas described above. Namely draw $1, 1/2, 1/3,$ etc (of course it would not be possible to draw all of them!!).
Now an open ball in the metric space $\mathbb{R}$ with the usual Euclidean metric is just an open interval of the form $(-a,a)$ where $a\in \mathbb{R}$. Now we claim that $0$ is a limit point. How?
Given me an open interval about $0$. For now let it be $(-0.5343, 0.5343)$, a random interval I plucked out of the air. The question now is does this interval contain a point $p$ of the set $\{\frac{1}{n}\}_{n=1}^{\infty}$ different from $0$? Well sure, because by the archimedean property of the reals given any $\epsilon > 0$, we can find $n \in N$ such that
$$0 < \frac{1}{n} < \epsilon.$$
In fact you should be able to see from this immediately that whether or not I picked the open interval $(-0.5343,0.5343)$, $(-\sqrt{2},\sqrt{2})$ or any open interval.
Now let us look at the set $\mathbb{Z}$ as a subset of the reals. What you do now is get a paper, draw the number line and draw some dots on there to represent the integers. Can you see why you are able to draw a ball around an integer that does not contain any other integer?
Having understood this, looks at the following definition below:
$\textbf{Definition:}$ Let $E \subset X$ a metric space. We say that $p$ is a limit point of $E$ if for all $\epsilon > 0$, $B_{\epsilon} (p)$ contains a point of $E$ different from $p$.
$\textbf{The negation:} $ A point $p$ is not a limit point of $E$ if there exists some $\epsilon > 0$ such that $B_{\epsilon} (p)$ contains no point of $E$ different from $p$.
From the negation above, can you see now why every point of $\mathbb{Z}$ satisfies the negation? You already know that you are able to draw a ball around an integer that does not contain any other integer.
Best Answer
HINTS for (1) and (2) and answers to (3) and (4):
Every non-empty open set in $\Bbb R$ is a union of open intervals. Does $\Bbb Q$ contain any non-empty open interval?
What is $\operatorname{cl}\varnothing$?
$\operatorname{int}\operatorname{cl}A=\varnothing$ if and only if $\operatorname{cl}A$ contains no non-empty open set. An example of such an $A$ is $\left\{\frac1n:n\in\Bbb Z^+\right\}$: the closure of this set is $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, which contains no non-empty open interval. Another way to say this: every point of $\operatorname{cl}A$ is a limit point of $\Bbb R\setminus\operatorname{cl}A$, the complement of $\operatorname{cl}A$.
If $A\subseteq\Bbb R$, a point $x\in\Bbb R$ is a limit point of $A$ if and only if every open neighborhood of $x$ contains a point of $A\setminus\{x\}$. If you’re working in $\Bbb R$, as you seem to be, real numbers are the only things that can be limit points of $A$: they’re the only things that are even points in your space.