a)When A and B are independent events P(A and B) = P(A) * P(B)

let denote the event wining the game with A

A is independent event

thus we have

P(AAA) = P(A and A and A) = P(A)P(A)P(A) = 0.7^3 = 0.343

the probability of Jack winning all 3 games is 0.343

b)lets B be the event that Jack has lost the game thus

P(B) = 1 - P(A) = 1 - 0.7 = 0.3

Jack winning exactly 2 games will be the following events

AAB - Jack winning the first 2 games and losing the third

ABA - Jack winning the first and the third game and losing the second

BAA - Jack winning the last 2 games and losing the first

P(AAB or ABA or BAA) = P(AAB) + P(ABA) + (BAA) = P(A)P(A)P(B) + P(A)P(B)P(A) + P(B)P(A)P(A) = 3P(A)P(A)P(B) = 3*0.7*0.7*0.3

P(AAB or ABA or BAA) = 0.441

the probability of Jack winning exactly 2 games is 0.441.