6.4: A number is divisible by both $3$ and $5$ if it is divisible by $3 \times 5 = 15$. Since $99 \div 5 = 19.8$, there are $19$ multiples of $5$ less than $100$. Since $99 \div 15 = 6.6$, there are $6$ multiples of $15$ less than $100$.

6.8: Let $C$ be the set of adults who like coffee and $T$ the set of adults who like tea. Then $n(C) = 30$, $n(T) = 30$, and $n(C \cup T ) = 50$. Since $n(C \cup T) = n(C) + n(T) - n(C \cap T)$, we have $n(C \cap T) = 30 + 30 - 50 = 10$, so there are $10$ adults who like both coffee and tea.

6.9.(a): You can pretend Sarah throws a coin to decide whether or not to eat salad and another coin to decide whether or not to eat dessert. There are $4$ possible outcomes for the coins: $HH, HT, TH, TT$. 

6.9(b): For each of the $7$ days she has $4$ choices (the ones from part(a)), so the product rule tells us we have $\underbrace{4 \times \dots \times 4}_{7 \text{ times}} = 4^7$ different outcomes for the whole week.

6.24: To find the median of a list of $9$ numbers, we first order them from smallest to largest, then the median is the number in the middle of the list (since $9$ is odd). To make the difference as large as possible, we want to make the median as small as possible and the mean as large as possible (or vice versa). To have the median be equal to $0$, we need the first $5$ numbers to be $0$, and to have the mean as big as possible, the remaining $4$ numbers should be all $20$, so the mean is $\dfrac{0 + 0 + 0 + 0 + 0 + 20 + 20 + 20 + 20}{9} = \dfrac{80}{9}$.

6.30: The mean of four numbers $a$, $b$, $c$, $d$, is $\dfrac{a + b + c  + d}{4}$. Since the mean of the numbers is $71.5$, it must be true that $a + b + c + d = 71.5 \times 4 = 286$. We know three of the numbers, so we can use their sum to find the fourth number: $286 - (58 + 76 + 88) = 64$.