I'm struggling quite a bit with this question:
Prove that if $E \cap F = \emptyset $, then:
$ \textbf{P}(E|E \cup F) = \frac{\textbf{P}(E)}{\textbf{P}(E)+\textbf{P}(F)} $
Using $ A \cap (A \cup B) = (A \cap A) \cup (A \cap B) $:
${\textbf{P}(E|E \cup F) = \frac{\textbf{P}(E \cap (E \cup F))\textbf{P}(E)}{\textbf{P}(E \ cup F}}$
$= \frac{(\textbf{P}(E \cap E) + \textbf{P}(E \cap F))\textbf{P}(E)}{\textbf{P}(E) + \textbf{P}(F)}$
$= \frac{(\textbf{P}(E))^2}{\textbf{P}(E) + \textbf{P}(F)}$
On
We know that $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
Further we know that $E\cap(E\cup F) = E$. Now if $E\cap F = \emptyset$, then $$P(E|E\cup F) = \frac{P(E\cap(E\cup F))}{P(E\cup F}$$ \begin{align*} P(E|E\cup F) &= \frac{P(E\cap(E\cup F))}{P(E\cup F)}\\ &= \frac{P(E)}{P(E) + P(F) - P(E\cap F)}\\ &= \frac{P(E)}{P(E) + P(F)} \end{align*}
I expect that what you did in your attempt was first $Pr(E\mid E\cup F) = \dfrac{Pr(E\cap (E\cup F))}{Pr(E\cup F)}$ and then you broke things apart with "addition and multiplication rules" but you went and applied this rules incorrectly.
You should know the following:
$Pr(A\cap B) = Pr(A)\times Pr(B)$ is NOT true in general. This is true if and only if $A$ and $B$ are independent events.
$Pr(A\cup B) = Pr(A)+Pr(B)$ is NOT true in general. This is true if $A$ and $B$ are mutually exclusive events and only if $A\cap B$ is an impossible or an almost impossible event.
Instead, the correct statements are:
$Pr(A\cap B) = Pr(A)\times Pr(B\mid A)$
$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$
You went and did $\dfrac{Pr(E\cap (E\cup F))}{Pr(E\cup F)} = \dfrac{Pr((E\cap E)\cup (E\cap F))}{Pr(E\cup F)} = \dfrac{Pr(E\cap E)+Pr(E\cap F)}{Pr(E)+Pr(F)} = \dfrac{Pr(E)\times Pr(E)+Pr(\emptyset)}{Pr(E)+Pr(F)}$
You went and misused each of the mentioned properties when they shouldn't have been used.
Instead, simply note that $E\cap (E\cup F) = E$ and note that $E$ and $F$ are mutually exclusive, so we have $Pr(E\mid F) = \dfrac{Pr(E\cap (E\cup F))}{Pr(E\cup F)}=\dfrac{Pr(E)}{Pr(E)+Pr(F)}$