I think you misread the meaning of $(\frac1{n+1},\frac1n)$: It is the open interval with endpoints $\frac1{n+1}$ and $\frac1n$.
The set is the union of open sets and hence open, which implies that all points of $X$ are interior. Indeed, if $x\in X$, then $x\in(\frac1{n+1},\frac1n)$ for some $n$, i.e. $\frac1{n+1}<x<\frac1n$. With $\epsilon=\min\{x-\frac1{n+1},\frac1n-x\}>0$ the $\epsilon$ ball around $x$ is in $X$.
The set of cluster point and the closure are both equal to $[0,1]$ as precisely the $x$ with $0\le x\le 1$ have at least one/infintely many other points in every open neighbourhood.
The boundary $\partial X$ is the difference of closure and interior, i.e. $\{\frac1n\mid n\in\mathbb N\}\cup\{0\}$.
A limit point is a point of a set $S$, is a point $x$, which may or may not be an element of the set $S$, such that for every possible real number $\epsilon > 0$. There will exist an element $y \in S$, $y \ne x$ such that the distance between $x$ and $y$ is less than $\epsilon$.
In set $A$, $1$ is a limit point because for every $\epsilon > 0$ I can find an even $n$ so that $0 < 2/n < \epsilon$ (by finding and even $n > 2/\epsilon$) So $(-1)^n + 2/n = 1 + 2/n \in A$ and the distance between $1$ and $1 + 2/n$ is $2/n < \epsilon$. I can find such an $n$ for every possible epsilon.
$-1$ is also a limit point because for every $\epsilon > 0$ I can find an odd $n$ so the $0 < 2/n < \epsilon$ (by finding an odd $n > 2/\epsilon$) so $(-1)^n + 2/n = -1 + 2/n \in A$ and the distance between $-1$ and $-1 + 2/n$ is $2/n < \epsilon$.
Furthermore these are the only two limit points. This is a bit harder to show. Basically if $x$ is between $(-1)^{n-1}+2/(n-1)$ and $(-1)^{n+1} + 2/(n+1)$ inclusively there are no other points of $A$ in that interval and we can find a very small distance that has no other points in it. If $x$ is out of that range we can find a distance with no points in that range.
For set $B$, $0$, $1$ and every value in between are limit points. For any distance $\epsilon > 0$ and any $x \in [0,1]$ we can find rational numbers in $(x-\epsilon, x + \epsilon)$ that are between $0$ and $1$ exclusively. $[0,1]$ are the only numbers where that is true.
How do they relate to closed sets? The definition of a closed set is that all the limit points are in $S$. As $1 \not \in A$, $-1 \not \in A$, $0 \not \in B$,, and $1 \not \in B$, neither $A$ nor $B$ are closed.
How do you find the isolated points? Okay, all the points in A are isolated and none of the points in B are. An isolate point of $S$, is a point $x \in X$ is one in which there exists a distance $\epsilon > 0$ so that there are no other point of $S$ within that distance from $x$. This is the exact opposite of a limit point. In $A$ all points are isolated and are not limit points. In $B$, none of the points are isolated as far any distance from any point, there will be a rational number in the set. So no point in $B$ is isolated and in fact all points in $B$ are limit points of $B$. (But $B$ is not closed because there are other limit points that are not in $B$).
So what is a closure? The closure of $S$ is the smallest closed set containing $S$. In other words $\overline{S}$ is the union of $S$ with its limit points. (Because to be closed means that it contains its limit points.
So $\overline{A} = A \cup $ limit points of $A = A \cup {0} \cup {1}$.
$\overline{B} = B \cup $ limit points of $B = \{x \in \mathbb Q| 0< x < 1\} \cup [0,1] = [0,1]$. (Note. This includes all the irrationals as well as just the rationals. The irrationals between 0 and 1 are limit points as well.)
Best Answer
You are correct. I have one little comment. Since you have already ascertained that $-1$ does not belong to $A$, the phrase "except -1" in (c) is not needed.