(A) The sample space $S$ for tossing three coins is:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$,so $n(S) = 8$.
The events are:
$E = \{HHH, TTT\} \implies P(E) = \frac{2}{8} = \frac{1}{4}$
$F = \{HHH, HHT, HTH, THH\} \implies P(F) = \frac{4}{8} = \frac{1}{2}$
$G = \{HHT, HTH, THH, HTT, THT, TTH, TTT\} \implies P(G) = \frac{7}{8}$
Intersections:
$E \cap F = \{HHH\} \implies P(E \cap F) = \frac{1}{8}$
$E \cap G = \{TTT\} \implies P(E \cap G) = \frac{1}{8}$
$F \cap G = \{HHT, HTH, THH\} \implies P(F \cap G) = \frac{3}{8}$
Checking independence $(P(A \cap B) = P(A) \cdot P(B))$:
$1$. For $(E, F)$: $P(E) \cdot P(F) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} = P(E \cap F)$. Thus,$(E, F)$ are independent.
$2$. For $(E, G)$: $P(E) \cdot P(G) = \frac{1}{4} \cdot \frac{7}{8} = \frac{7}{32} \neq P(E \cap G) = \frac{1}{8}$. Thus,$(E, G)$ are dependent.
$3$. For $(F, G)$: $P(F) \cdot P(G) = \frac{1}{2} \cdot \frac{7}{8} = \frac{7}{16} \neq P(F \cap G) = \frac{3}{8}$. Thus,$(F, G)$ are dependent.