Conditional probability homework help

Problem 1:
A college student is taking two courses. The probability she passes the first course is 0.76. The probability she passes the second course is 0.68. The probability she passes at least one of the courses is 0.95.
1) What is the probability she passes exactly one course?
2) Given she passes the first course, what is the probability she does not pass the second?

Problem 2:
The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags. Based on many test runs, the TSA determines that if a bag does contain large amounts of liquid, there is a probability of 0.9 the test will detect it. If a bag does not contain large amounts of liquid, there is a 0.1 probability the test will conclude that it does (a false positive). Suppose that in reality only 5 in 100 bags actually contain large amounts of liquid.
1) What is the probability a randomly selected bag will have a positive test?
2) Given a randomly selected bag has a positive test, what is the probability it actually contains a large amount of liquid?
3) Given a randomly selected bag has a positive test, what is the probability it does not contain a large amount of liquid?

I'd appriciate any help. Thanks

q1
P(A) = 0.76, P(B) = 0.68, P(A or B) = 0.95

P(A or B) = P(A) + P(B) - P(AB)
0.95 = 0.76 + 0.68 - P(AB)
so P(AB) = 0.76+0.68-0.95 = 0.49

1. P(A or B) - P(AB) = 0.95 - 0.49 = 0.46 <------

2. P(AB') = 0.76 - 0.49 = 0.27, P(AB')/P(A) = 0.27/0.76 = 0.3553

draw a venn diagm. for better understanding

q2
P[true positive] = 0.9*0.05 = T
Pfalse positive] = 0.1*0.95 = F

a. P[positive] = T + F = 0.14 <-----

b. P[ T | positive] = T/0.14 = 0.3214 <------

c. P[ F | positive] = 1 - 0.3214 = 0.6786 <-------