(A) Total possible outcomes of tossing a coin $3$ times:
$S = \{(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)\}$
$\therefore n(S) = 8$
$(i)$ Let $E_1$ be the event that Sweta loses the entry fee.
This happens if she gets zero heads (i.e.,$TTT$).
$E_1 = \{(TTT)\}$,so $n(E_1) = 1$.
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{8}$.
$(ii)$ Let $E_2$ be the event that Sweta gets double the entry fee.
This happens if she gets $3$ heads (i.e.,$HHH$).
$E_2 = \{(HHH)\}$,so $n(E_2) = 1$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{8}$.
$(iii)$ Let $E_3$ be the event that Sweta gets her entry fee back.
This happens if she gets one or two heads.
$E_3 = \{(HTT), (THT), (TTH), (HHT), (HTH), (THH)\}$,so $n(E_3) = 6$.
$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{6}{8} = \frac{3}{4}$.