Probability that Rebecca will be chosen to represent her school is $\frac{\binom{4}{1}}{\binom{8}{1}} = \frac{4}{8}$
Probability that Elise will be chosen is $\frac{\binom{4}{1}}{\binom{9}{1}} = \frac{4}{9}$
Consider X as school 1 and Y as school 2, Rebecca in school 1 and Elise in school 2.
(a) So here Rebecca have equally likely chances to pair with all 4 opponents. Therefore for each opponent in school 2 she have $\frac{1}{4}$ chances that means probability that they (Rebecca & Elise) will be paired is $\frac{1}{4}$.
Now finally coming to the probability that Rebecca & Elise will be paired
$\Rightarrow \frac{4}{8} * \frac{4}{9} * \frac{1}{4} \Rightarrow \frac{1}{18}$
(b) We know to play with each other they have they have 1/4 chances. So probability of not to play with each other they have $1-\frac{1}{4}= \frac{3}{4}$ (or it can be observed with diagram)
Therefore probability that they will be chosen but will not play each other$\Rightarrow \frac{4}{8} * \frac{4}{9} * \frac{3}{4}\Rightarrow \frac{1}{6}$
(c) remember one thing when it is asked for EITHER OR CASE (EXACTLY ONE OF) then it becomes the case of EXCLUSIVE OR (XOR) "where it is one or the other but not the both"
this becomes $P(A\cap {B}') \cup P({A}'\cap B)$
$\Rightarrow (\frac{4}{8}*\frac{5}{9}) + (\frac{4}{8}*\frac{4}{9})$
$\Rightarrow \frac{20}{72}+\frac{16}{72}= \frac{36}{72} \Rightarrow \frac{1}{2}$
