Question 1:
(a) [Lower Limit Topology] You are correct that the closures of $(0,\sqrt{2})$ and $(\sqrt{2},3)$ in the lower limit topology are $[0,\sqrt{2})$ and $[\sqrt{2},3)$, respectively.
(b) [Other Topology] Hint: Prove that the closure of $(0,\sqrt{2})$ in this topology is $[0,\sqrt{2}]$; use the irrationality of $\sqrt{2}$. Can you see that the closure of $(\sqrt{2},3)$ in this topology is $[\sqrt{2},3)$?
Question 2:
Let us choose the "standard basis" for the product topology on $B\times D$ consisting of all sets of the form $U\times V$ where $U$ is open in $B$ and $V$ is open in $D$. We wish to establish the continuity of $f\times g$.
(a) Show that it suffices to prove the preimage under $f\times g$ of each element of this basis is open in $A\times C$. (Hint: use the definition of a basis.)
(b) Prove that if $U$ is open in $B$ and $V$ is open in $D$, then $(f\times g)^{-1}(U\times V)=f^{-1}(U)\times g^{-1}(V)$. Deduce that $f\times g$ is a continuous function.
$\mathcal{T}_1:$ 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.
Best Answer
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.)