Since $\text{R}$ is reflexive so $\text{R} \cup \text{S}$ will also be reflexive. Since $\text{R, S}$ are symmetric, so $\text{R} \cup \text{S}$ also will be symmetric. But $\text{R} \cup \text{S}$ may not be transitive.
Consider this $: \text{A} = \{1,2,3\} , \text{R} = \{(1,1), (2,2), (3,3), (1,2),(2,1)\}$ and $\text{S} = \{(3,2), (2,3),(2,2),(3,3)\}$
$\text{R} \cup \text{S}$ is Not transitive.