(b) Given, $R$ and $S$ are relations on set $A$.

 $\therefore R \subseteq A \times A$and $S \subseteq A \times A$ ==> $R \cap C \subseteq A \times A$

==> $R \cap S$is also a relation on $A$.

Reflexivity : Let a be an arbitrary element of $A$. Then, $a \in A$==> $(a,a) \in R$and $(a,a) \in S$,

[$\therefore$ $R$ and $S$ are reflexive]

==> $(a,\,a) \in R \cap S$

Thus, $(a,a) \in R \cap S$for all $a \in A$.

So, $R \cap S$is a reflexive relation on $ A$.

Symmetry : Let $a,b \in A$such that $(a,b) \in R \cap S$.

Then, $(a,b) \in R \cap S$ ==> $(a,b) \in R$and $(a,b) \in S$

==> $(b,a) \in R$ and $(b,a) \in S$,

[ $\therefore $ $R$ and $S$ are symmetric]

==> $(b,a) \in R \cap S$

Thus, $(a,b) \in R \cap S$

==> $(b,a) \in R \cap S$ for all $(a,b) \in R \cap S$.

So, $R \cap S$ is symmetric on $A$.

Transitivity : Let $a,b,c \in A$ such that $(a,b) \in R \cap S$ and $(b,c) \in R \cap S$. 

Then, $(a,b) \in R \cap S$ and $(b,c) \in R \cap S$

==> $\{ ((a,b) \in R\,\,{\rm{and}}\,(a,b) \in S\,)\} $

and $\{ ((b,c) \in R\,{\rm{and}}\,{\rm{(}}b,c{\rm{)}} \in S\} $

==> $\{ (a,b) \in R,(b,c) \in R\} $ and $\{ (a,b) \in S,(b,c) \in S\} $

==> $(a,c) \in R$ and $(a,c) \in S$

$\left[ \begin{gathered}
  \because R\,{\text{and}}\,S\,{\text{are}}\,{\text{transitive}}\,{\text{So}} \hfill \\
  {\text{(}}a,b{\text{)}} \in R\,{\text{and (}}b,c{\text{)}} \in R \Rightarrow {\text{(}}a{\text{,}}c{\text{)}} \in R \hfill \\
  {\text{(}}a,b{\text{)}} \in S{\text{ and (}}b{\text{,}}c{\text{)}} \in S \Rightarrow {\text{(}}a{\text{,}}c{\text{)}} \in S \hfill \\ 
\end{gathered}  \right.$

==> $(a,c) \in R \cap S$

Thus,$(a,b) \in R \cap S$ and$(b,c) \in R \cap S \Rightarrow (a,c) \in R \cap S$.

So, $R \cap S$ is transitive on $A$.

Hence, $R$ is an equivalence relation on $A$.