#1 is simply a case of finding P(X > 130). If you are using probability tables then change to the Z variable:
#2: P(T < k) = 0.1. You need to find a value k so that 10% of gestation periods are shorter than the time k. Inverse normal distribution.
Change to standard variable Z: .
Looking at a probability table we can see that P(0 < Z < 1.282) = 0.4. This means that P(0 > Z > -1.282) = 0.4, as the normal distribution is symmetric about the mean.
So combining P(0 > Z > -1.282) = 0.4 and P(Z > 0) = 0.5 we get P(Z > -1.282) = 0.9 and thus P(Z < -1.282) = 0.1.
So the value . Now solve for k to find the answer.