Your example is correct, and you don't need to do any more work! To show a relation is not an equivalence relation, you only need to show that one of the three conditions (reflexive, symmetric, transitive) fails. To be an equivalence relation, all three need to be true, as as soon as one of them fails, it's not an equivalence relation. So you have shown that $R_1\circ R_2$ is not an equivalence relation in your example.
(In fact, if $R_1$ and $R_2$ are reflexive, then $R_1\circ R_2$ always is reflexive as well. The proof is easy: for any $x\in A$, $(x,x)\in R_1$ and $(x,x)\in R_2$, so $(x,x)\in R_1\circ R_2$ (taking the $y$ in the definition of $R_1\circ R_2$ to be $x$).)
Hint: In c) one requires to enumerate all partitions of the set $X$ (up to isomorphism). First calculate the number of partitions of $X$. This is the Bell number $B(5)$, where $B(n) = \sum_{k=1}^n S(n,k)$ and $S(n,k)$ is the number of partitions of $n$ into $k$ parts (Stirling number of the 2nd kind).
We have $B(1)=1$, with partition $\{\{1\}\}$,
$B(2)=2$ with partitions $\{\{1,2\}\}$, $\{\{1\},\{2\}\}$,
$B(3)=5$ with partitions $\{\{1\},\{\{2\},\{3\}\}$, $\{\{1,2\},\{3\}\}$, $\{\{1,3\},\{2\}\}$, $\{\{2,3\},\{1\}\}$, $\{\{1,2,3\}\}$, and so on.
In view of the isomorphism classes, you just need to consider the types of partitions (see comment below).
BY REQUEST: For instance, take the bijection $f=\left(\begin{array}{ccc}1&2&3\\2&3&1\end{array}\right)$. Then the partition $\{\{1,2\},\{3\}\}$ is mapped to the partition $\{\{f(1),f(2)\},\{f(3)\}\} = \{\{2,3\},\{1\}\}$.
Best Answer
We use the fact that for every reflexive binary relations $R, S$ on a set, $R \subseteq R \circ S$ and $R \subseteq S \circ R$, and that $R$ is transitive iff $R \circ R = R$.
$(\Rightarrow:)$ Let us suppose that $R \circ S$ is transitive and let $(a,c) \in R \circ S$.
Then, for some $b$, we have $(a,b) \in R$ and $(b,c) \in S$.
Since the relations are symmetric, $(c,b) \in S$ and $(b,a) \in R$, from which it follows by an earlier remark that $(c,b) \in R \circ S$ and $(b,a) \in R \circ S$.
Now, by hypothesis, it follows that $(c,a) \in R \circ S$, whence $(a,c) \in S \circ R$.
We conclude that $R \circ S \subseteq S \circ R$.
The reverse inclusion follows from $S \circ R \subseteq R \circ S \circ R \circ S = R \circ S$, by hypothesis.
So $R \circ S = S \circ R$.
$(\Leftarrow:)$ Suppose $(a,b), (b,c) \in R \circ S$.
Then, there exist $u,v$ such that $$a R u S b \quad \text{ and } \quad b R v S c.$$ From $b R v S c$ it follows that $(b,c) \in R \circ S = S \circ R$, and so $(b,c) \in S \circ R$.
Now, from $u S b (S \circ R) c$ it follows that $(u,c) \in S \circ R$, and from $a R u (S \circ R)c$, it follows that $(a,c) \in R \circ (S \circ R) = R \circ (R \circ S) = R \circ S$, and so $R \circ S$ is transitive.