Let $k$ be the set of your question in your notation:
Hint for (1) [standard topology]: Prove that the point $0$ is in the closure of $k$. If $x<0$, then $x$ cannot be in the closure of $k$ since $(x-1,0)$ is a neighborhood of $x$ that does not intersect $k$. Similarly, prove that if $x>1$, then $x$ is not in the closure of $k$. Of course, every point of $k$ is in the closure of $k$. However, if $0<x<1$ is a point not in $k$, prove that $x$ is strictly between two points of $k$; that is, there is an integer $n\geq 1$ such that $\frac{1}{n+1}<x<\frac{1}{n}$. Deduce that $x$ is not in the closure of $k$. Can you now tell what the closure of $k$ is?
Hint for (2) [finite complement topology]: Note that $k$ is infinite and that the complement of any non-empty open set in this topology is finite. Can you now tell what the closure of $k$ is?
Hint for (3) [lower limit topology]: The closure of $k$ in this topology is exactly the same as in (1). Why?
The following exercises (with hints) will further your understanding of this problem:
Exercise 1:
Let $T$ be the upper limit topology on $\mathbb{R}$; that is, $T$ is the topology generated by the base of all sets of the form $(a,b]$ where $a<b$. What is the closure of $k$ in the topology $T$. (Hint: The real question is whether or not $0$ is in the closure of $k$.)
Exercise 2:
What is the closure of $S=\{1-\frac{1}{n}:n\geq 1\}$ in the lower limit topology? (Hint: Prove that $1$ is not in the closure of $S$.) What is the closure of $S$ in the upper limit topology?
Exercise 3:
The cocountable topology on $\mathbb{R}$ is the topology consisting of $\emptyset$ and all subsets of $\mathbb{R}$ whose complement is countable. What is the closure of $k$ (the set of your question) in the cocountable topology on $\mathbb{R}$? What is the closure of $(0,1)=\{x:0<x<1\}$ in the cocountable topology on $\mathbb{R}$? (Hint: $k$ is countable and $(0,1)$ is uncountable. Therefore, the complement of $k$ is open in the cocountable topology. However, the complement of $(0,1)$ is not open in the cocountable topology.)
Observation: for any two point set $X$ the set of topologies on $X$ has $4$ members (discrete, indiscrete and two Sierpinski variants) and has the property that any pair of them has a union that is a topology again. So the set of topologies on $X$ is closed under (all/finite) unions.
For a three point set $X$ you can find incompatible topologies whose union is no longer a topology like $\{0\}$ (as the only non-trivial open set on $X=\{0,1,2\}$) for $\mathcal{T}_1$ and for $\mathcal{T}_2$ the topology with non-trivial open sets $\{1\}, \{2\}, \{1,2\}$: the union has both $\{0\}$ and $\{1\}$ in it but not $\{0,1\}$ so is not a topology. This example also shows the same for any larger set, of course, using only finite topologies like these...
So the only sets on which topologies are closed under finite unions are those of size at most $2$...
Best Answer
$\mathcal{T}_1:$ That's right, well done. Note that this topology is generated by the intervals $(a,b)$. So for any point $x$ in $\mathbb{R}$ in between $k = \frac1n$ and $k^\prime = \frac1m$ we can choose $(x - m, x + m)$ where $m = \frac12 \min (\frac1n , \frac1m)$ to see that $x$ is not in the closure of $K$. Looking at $0$ we see that every interval $0 \in (a,b)$ has to have $0 < b$ so there will be an $n$ such that $0 < \frac1n < b$ and hence every open set containing $0$ will have non-empty intersection with $K$. Hence we get $\overline{K} = K \cup \{0\}$.
$\mathcal{T}_2$: Note that $\mathcal{T}_2$ is generated by $(a,b) $ and $ (a,b) \setminus K $ so $\mathcal{T}_1 \subset \mathcal{T}_2$ and hence $ \overline{K}_{\mathcal{T}_2} \subset \overline{K}_{\mathcal{T}_1}$ since we have more sets that might contain our point but might not intersect with $K$. Since $K \subset \overline{K}$ we therefore know that the only point we need to think about is $0$. But none of the sets we added to the basis intersect with $K$ hence there now are open sets $O$ containing $0$ with $O \cap K = \varnothing$. Hence $0$ is not in the closure of $K$ and we have $\overline{K}=K$.
$\mathcal{T}_3:$ Note that the open sets in this topology look like this: $O = \mathbb{R} \setminus \{x_1, \dots, x_n \}$. Consider a point $p$ in $\mathbb{R}$. Let $O$ be an open set containing $p$. Note that $O$ has non-empty intersection with $K$ since $O$ contains all points of $\mathbb{R}$ except finitely many. In particular, $K$ contains infinitely many points hence their intersection will be non-empty. Since $p$ was arbitrary we get $\overline{K} = \mathbb{R}$.
$\mathcal{T}_4$: Note that the upper limit topology is generated by intervals of the form $(a, b]$. Note that for a point $p \neq 0$ outside $K$ we can pick $(\varepsilon , p]$ with $\varepsilon$ small enough such that $(\varepsilon , p] \cap K = \varnothing$. For $0$ we pick $(a, 0]$ to get a set that does not intersect with $K$ hence $0$ is not in the closure of $K$. Hence we get $\overline{K} = K$.
$\mathcal{T}_5$: Again you have it right, well done. To see this pick a point $p$ to the right of $0$ or $0$ itself and note that every open interval $(-\infty, a)$ containing $p$ will intersect with $K$ since there will be an $n$ such that $\frac1n < p$.
Hope this helps.