Answer:
(a) F(A, B, C) = AB'C' + A'BC' + A'B'C
(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)
Step-by-step explanation:
(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either
A is heads and the others are tails (AB'C')
B is heads and the others are tails (A'BC')
C is heads and the others are tails (A'B'C)
Hence, as a minterm expansion,
F(A, B, C) = AB'C' + A'BC' + A'B'C
(b) To get the corresponding maxterm expansion, we convert to binary.

The maxterm is the product of the complements.

Expanding,
F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)