$L \in RE/R$ such that $L^R \cup L \in R$
Since this sounds like a homework question to me, I will give the answer as a series of hints. (You should make a serious effort to solve the question after each hint, instead of just reading them all in short succession.)

First: Such an $L$ exists and $L \cup L^R$ is not necessarily $\Sigma^*$.


   Try $L \cup L^R = \{a^nb^n\} \cup \{b^na^n\}$.





   For each $n$, $L$ should contain exactly one of the words $a^nb^n$, $b^na^n$. How can you do this so that $L$ is in $RE/R$?


And finally the solution:


   Let $K$ be any set in $RE/R$ and let $L = \{a^nb^n \mid n \in K\} \cup \{b^na^n \mid n \notin K\}$.