= Solution
Use the convention that $R\circ S$ applies $S$ first. Reflexivity inserts an identity step: if $aRb$, use $aSa$ to obtain $(a,b)\in R\circ S$; if $aSb$, use $bRb$ for the same conclusion. Thus $R\cup S\subseteq R\circ S$. By <monotonicity of relation composition>,
$$
(R\cup S)\circ(R\cup S)\subseteq(R\circ S)\circ(R\circ S).
$$
For strictness, take $A=\{0,1,2,3\}$ and let $R=S$ consist of the identity pairs together with $(0,1),(1,2),(2,3)$. Both are <reflexive relations>. The left side permits at most two forward chain steps, so does not contain $(0,3)$. The right side permits four steps, including $0\to1\to2\to3\to3$, so does contain it. Equality therefore need not hold.
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