(Assuming that two tests are independent, which is a rather unrealistic assumption in this case) If you know how to calculate the ~30% answer to the first part of the question, then this problem is pretty straightforward to solve. Just use Bayes' rule again, treating the posterior from your first calculation (~30%) as your prior for the next calculation. If Kim came from a population that had a ~30% prior of having cancer and took one test which came out negative, then her probability after that one test would be the same as Jane's probability after both tests.

Doing this with probabilities is a bit more complicated than what Coscott did, but to illustrate it anyhow... where A is cancer and and C are the two test results, P(A|BC)=P(A) P(BC|A) / P(BC). P(A) is our prior of 5%. Because B and C are independent, P(BC|A) is just 0.8 * 0.2. P(BC) is where using probabilities is more complicated than using odds, because it's not the probability of false positives, it's the total prior probability of seeing B and then C. Using the product rule, P(BC) = P(B)*P(C|B). Then splitting the possibilities up into cancer and not-cancer, this becomes (P(AB)+P(¬A B))*(P(AC|B)+P(¬A C|B)). Because B and C are independent, the second part becomes (product rule) P(A|B)*P(C|A)+P(¬A|B)*P(C|¬A) - note that even when we added both pieces of evidence at once, we still have to calculate the intermediate probability P(A|B)! Stupid non-time-saving grumble grumble. Anyhow, if you plug in the numbers, it's ~0.094.