Let $R=\{\langle 1,2\rangle,\langle 2,1\rangle,\langle 2,3\rangle,\langle 3,4\rangle,\langle 4,1\rangle\}$ on $\{1,2,3,4\}$. What failure of transitivity do we have here?
$$\begin{align*}
&\langle 1,2\rangle,\langle 2,1\rangle\in R,\text{ but }\langle 1,1\rangle\notin R\\
&\langle 1,2\rangle,\langle 2,3\rangle\in R,\text{ but }\langle 1,3\rangle\notin R\\
&\langle 2,1\rangle,\langle 1,2\rangle\in R,\text{ but }\langle 2,2\rangle\notin R\\
&\langle 2,3\rangle,\langle 3,4\rangle\in R,\text{ but }\langle 2,4\rangle\notin R\\
&\langle 3,4\rangle,\langle 4,1\rangle\in R,\text{ but }\langle 3,1\rangle\notin R\\
&\langle 4,1\rangle,\langle 1,2\rangle\in R,\text{ but }\langle 4,2\rangle\notin R\\
\end{align*}$$
This means that in order to expand $R$ to a transitive relation, we must add at least the six ordered pairs $\langle 1,1\rangle,\langle 1,3\rangle,\langle 2,2\rangle,\langle 2,4\rangle,\langle 3,1\rangle$, and $\langle 4,2\rangle$. This gives us the relation
$$R\,'=\{\langle 1,1\rangle,\langle 1,2\rangle,\langle 1,3\rangle,\langle 2,1\rangle,\langle 2,2\rangle,\langle 2,3\rangle,\langle 2,4\rangle,\langle 3,1\rangle,\langle 3,4\rangle,\langle 4,1\rangle,\langle 4,2\rangle\}\;.$$
Are there any failures of transitivity here?
$$\begin{align*}
&\langle 1,2\rangle,\langle 2,4\rangle\in R\,',\text{ but }\langle 1,4\rangle\notin R\,'\\
&\langle 1,3\rangle,\langle 3,4\rangle\in R\,',\text{ but }\langle 1,4\rangle\notin R\,'\\
&\langle 3,1\rangle,\langle 1,3\rangle\in R\,',\text{ but }\langle 3,3\rangle\notin R\,'\\
&\langle 3,1\rangle,\langle 1,2\rangle\in R\,',\text{ but }\langle 3,2\rangle\notin R\,'\\
&\langle 3,4\rangle,\langle 4,2\rangle\in R\,',\text{ but }\langle 3,2\rangle\notin R\,'
\end{align*}$$
Thus, $R\,'$ isn’t yet transitive: we need to add at least the pairs $\langle 1,4\rangle,\langle 3,2\rangle$, and $\langle 3,3\rangle$ to $R\,'$ to have any hope of having a transitive relation. This gives us the relation
$$R''=\{\langle 1,1\rangle,\langle 1,2\rangle,\langle 1,3\rangle,\langle 1,4\rangle,\langle 2,1\rangle,\langle 2,2\rangle,\langle 2,3\rangle,\langle 2,4\rangle,\langle 3,1\rangle,\langle 3,2\rangle,\langle 3,3\rangle,\langle 3,4\rangle,\langle 4,1\rangle,\langle 4,2\rangle\}\;.$$
Note that in $R''$ the elements $1,2$, and $3$ of $A$ are all related to everything in $A$; only $4$ is not. And that gives us some failures of transitivity in $R''$:
$$\begin{align*}
&\langle 4,1\rangle,\langle 1,3\rangle\in R'',\text{ but }\langle 4,2\rangle\notin R''\\
&\langle 4,1\rangle,\langle 1,4\rangle\in R'',\text{ but }\langle 4,4\rangle\notin R''
\end{align*}$$
There are some other failures, but in each case the missing pair is either $\langle 4,3\rangle$ or $\langle 4,4\rangle$. Thus, to make $R''$ transitive we must at least add these two pairs. Call the resulting relation $\overline{R}$. We can’t add any more pairs, because $\overline{R}=A\times A$: it already contains every possible ordered pair. And finally we do have a transitive relation.
Because at each stage I added only those ordered pairs that that I had actually shown to be necessary for transitivity, this $\overline{R}$ is the smallest transitive relation containing the original relation $R$; by definition it is the transitive closure of $R$. The problem asks you to find and write out the transitive closures of the other three relations as well.
There are much more efficient ways to find the transitive closure of a relation, but at this point it’s probably most instructive for you to do everything by hand, so to speak, as I did above.
Best Answer
As was noted in the comments, you have far too much in your second level. For instance, $2\mid 4$ and $1\mid 1$, so $\langle 2,1\rangle$ must be below $\langle 4,1\rangle$. On the other hand, we know that there’s nothing between them, because there is no integer $n\{1,2,3,4\}$ such that $2\mid n$, $n\mid 4$, and $n$ is neither $2$ nor $4$.
Here are the bottom three levels; can you finish it from here?