Let $C$ be the set of people who drink coffee, and $T$ be the set of people who drink tea.
We are given the following information:
- $\(P(C)=0.80\)$ (80\% of people drink coffee)
- $\(P(T$ and $C)=0.10 \times P(C)=0.10 \times 0.80=0.08\)$ (10\% of coffee drinkers also drink tea)
- $\(P($ neither $)=0.10$ ( $10 \%\)$ drink neither coffee nor tea)

This means that the percentage of people who drink at least one of the two drinks is 1 $\(0.10=0.90\)$ or $\(90 \%\)$.

Now let's calculate the requested percentages:
(A) Coffee but not tea

This is the percentage of people who drink coffee minus those who drink both coffee and tea.
$\(P(C$ and $\operatorname{not} T)=P(C)-P(T$ and $C)=0.80-0.08=0.72\)$
So, $72 \%$ of the people in the party drink coffee but not tea.
(B) Tea but not coffee

We know that $\(P(C$ or $T)=P(C$ and $\operatorname{not} T)+P(T$ and $\operatorname{not} C)+P(C$ and $T)\)$.
We also know that $\(P(C$ or $T)=0.90\)$.
So, $\(0.90=0.72+P(T$ and $\operatorname{not} C)+0.08\)$.
$\(0.90=0.80+P(T$ and $\operatorname{not} C)\)$.
$\(P(T$ and $\operatorname{not} C)=0.90-0.80=0.10\)$.
So, 10\% of the people in the party drink tea but not coffee.
(C) Tea

This is the percentage of people who drink tea but not coffee, plus the percentage of people who drink both tea and coffee.
$\(P(T)=P(T$ and $\operatorname{not} C)+P(T$ and $C)=0.10+0.08=0.18$\).
So, $18 \%$ of the people in the party drink tea.