Let $\Sigma = \{a,b \}$.
Note that $s \in L(r)$ $\iff$ $s_k \in \Sigma$ for all $k$ and there exists $i such that $s_i=b$, $s_j=a$.
A moment's reflection shows that, without loss of generality, we may assume $s_i = s_{i+1} = \cdots = s_{j-1} = b$.
If $s \in L(r') \cup L(r'')$, then it is clear that $s$ has the form $s = \omega_1 b \omega_2 a \omega_3$, where $\omega_k \in \Sigma^*$, and so $s \in L(r)$. Hence $L(r') \cup L(r'') \subset L(r)$.
Now suppose $s \in L(r)$. From the above reflection, we see that $s$ has the form $s=\omega_1b b^* a \omega_2$, with $\omega_k \in \Sigma^*$. Hence $ s \in L(r')$. Now let $i$ be the first index such that $s_i = b$ and let $j$ be the last index such that $s_j = a$. (Since $s \in L(r)$ we know that such indices exist and that $i < j$.) Then $s$ has the form $a^* b \omega_1 a b^*$, with $\omega_1 \in \Sigma^*$. Hence it follows that $s \in L(r'')$.