To determine the probability that $E$ occurs before $F$, we can ignore
all the (independent) trials on which neither $E$ nor $F$ occurred,
that is, $(E\cup F)^c$ occurred, since we are going to repeat the
experiment until one of $E$ and $F$ does occur. So, look at the
trial of the experiment on which one of $E$ and $F$ has occurred
for the very first time.
We are given that on this trial, the event $E \cup F$ has occurred.
But, we don't yet know which of the two has occurred. So, given the
knowledge that $E \cup F$ has occurred, what is the conditional
probability that it was $E$ that occurred (and so $E$ occurred before $F$
since this is the first time we have seen either $E$ or $F$)?
$$P(E \mid (E \cup F)) = \frac{P(E(E \cup F))}{P(E \cup F)}
= \frac{P(E \cup EF)}{P(E) + P(F) - P(EF)}
= \frac{P(E)}{P(E)+P(F)}$$
since $P(EF) = P(\emptyset) = 0$.
Alternatively, let $G = (E\cup F)^c = E^c \cap F^c$ be the event that neither
$E$ nor $F$ occurs on a trial of the experiment. Note that
$P(G) = 1 - P(E) - P(F)$. Then, the event $E$ occurs
before $F$ if and only if one of the following compound events occurs:
$$
E, (G, E), (G, G, E), \ldots, (\underbrace{G, G, \ldots, G,}_{n-1} E), \ldots
$$
where $(\underbrace{G, G, \ldots, G,}_{n-1} E)$ means $n-1$ trials on which $G$
occurred and then $E$ occurred on the $n$-th trial. The desired probability
is thus
$$P(E ~\text{before}~ F) = P(E) + P(G)P(E) + [P(G)]^2P(E) + \cdots
= \frac{P(E)}{1 - P(G)} = \frac{P(E)}{P(E)+P(F)}.$$
Supposing, as is usual (e.g. here) that mutually exclusive means that $P(A \cap B) = 0$ we have that $P(A \land \lnot B) = P(A \setminus B) = P(A) - P(A \cap B) = P(A) = 0.74$.
The final one is then just $P(A)$ too. $A$ cannot occur at the same time as $B$.
Best Answer
Draw a Venn diagram. On the left, you have 28% that smoke cigarettes. On the right, you have the 7% that smoke cigars. The intersection is 5%. Simple arithmetic will tell you the portions of the circles that smoke one but not the other.