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
Firstly, in order to prove transitivity of some relation $R$, you need to show that $(x,y),(y,z)\in R$ implies $(x,z)\in R$.
The problem with your "proof" for the union is that you assume that $(x,y),(y,z)$ both belong to the same relation $R_i$. In that case, you indeed have $(x,z)\in R_i$ and hence $(x,z)\in R_3$. The problem occurs when you take $(x,y)\in R_1$ and $(y,z)\in R_2$. For example, if $R_1=\{(x,y)\in\mathbb N^2\mid x-y\text{ is divisible by 2}\}$ and $R_2=\{(x,y)\in\mathbb N^2\mid x-y\text{ is divisible by 3}\}$, both $R_i$ are transitive, but $R_1\cup R_2$ isn't: Consider $(0,2)\in R_1$ and $(2,5)\in R_2$. Do you think $(0,5)\in R_1\cup R_2$?
Your "counterexample" concerning the intersection isn't actually a counterexample! $R_3$ is indeed transitive. More precisely, we need to prove $$\forall x,y,z: (x,y),(y,z)\in R_3 \implies (x,z)\in R_3.$$
This implication is vacuously tue, that is, the antecedent (the part before $\implies$) is never satisfied, and thus the implication holds (here we use that the implications "false implies true" and "false implies false" are both correct).