For $R\cap S$ being symmetric,
Let $(x,y)\in R\cap S\implies (x,y)\in R $ and $(x,y)\in S\implies (y,x)\in R$ and $(y,x)\in S$ (as $R,S$are symmetric) $\implies (y,x)\in R\cap S$.
Thus, $R\cap S$ is symmetric.
For transitive part($R\cup S$),
Let $(x,y)$ and $(y,z)\in R\cup S\implies $
Three cases possible:
Case 1: If both $(x,y)$ and $(y,z)\in R\implies (x,z)\in R\implies (x,z)\in R\cup S\implies $ Transitivity holds for $R\cup S $ in this case.
Case 2: If both $(x,y)$ and $(y,z)\in S\implies (x,z)\in S\implies (x,z)\in R\cup S\implies $ Transitivity holds for $R\cup S $ in this case too.
Case 3: Case 1: If $(x,y)\in R$ and $(y,z)\in S$ does not$ \implies $ $(x,z)\in R$ or $ S$ which does not $\implies (x,z)\in R\cup S\implies $ Transitivity need not hold for $R\cup S $ in this case.
Thus, Considering all the three cases, $R\cup S$ need not be transitive.