I believe the elements of $A/R$ are $\{0,2\}$, $\{1\}$ and $A/R = \{\{0,2\},\{1\}\}$. Continuing with the second part of the question, I believe that that set comes out to be $$ \{ (\{0,2\}, 0), (\{0,2\}, 2), (\{1\}, 1) \}. $$ But, since I am unsure about the first part, the second part may not be correct.
The set of equivalence classes is correct.
Now you have to find the maps $f\colon A/R \to A$ such that $f(x)\in x$. A map $f\colon A/R\to A$ is determined by specifying the image of each element in the domain and there are nine of them.
If we set $e=\{0,2\}$ and $o=\{1\}$, we have some restrictions for a map to be in the set: we need $f(e)\in e$, so either $f(e)=0$ or $f(e)=2$; moreover $f(o)\in o$ is the same as $f(o)=1$. Thus we have just two maps satisfying the requirement.
If you want to describe the maps as sets of ordered pairs, the functions are $$ \{(e,0),(o,1)\}\qquad\text{and}\qquad\{(e,2),(o,1)\} $$ so your set is $$ \bigl\{ \{(\{0,2\},0),(\{1\},1)\},\{(\{0,2\},2),(\{1\},1)\}\bigr\} $$