Let A and B denote the events that the man wins contracts X and Y respectively.
Then P(A) = 0.5
P(A') = 1 - 0.5 = 0.5
P(B') = 0.3
P(B) = 1 - 0.3 = 0.7
(a) The probability that the man wins both contracts = $0.5\times 0.7=0.35$.
(b) The probability that the man wins exactly one of the contracts is $P(A)\times P(B^{\prime })+P(B)\times P(A^{\prime })$
= $0.5\times 0.3+0.7\times 0.5$
= $0.15+0.35$
= $0.50$
(c) Neither of the contracts (i.e not X, not Y) = $P(A^{\prime })\times P(B^{\prime })$
= $0.5\times 0.3$
= $0.15$