Regarding to Question 1, the probability that the lot was produced on line 1 is P(B)=0.2 since there are 5 production lines in total and the probability of choosing any production line is simply 1/5, i.e. 0.2.
For the conditional probability P(A|B), which means the probability of A given B occurs, in this case, it means the probability of a fuse was defective given the fuse was drawn from production line 1. Thus, according to the information given, the probability of a fuse was defective is 5%. Since there were three fuses in the lot, and the other two fuses passed the test, so the probability of that is 95%*95%. Therefore, the probability of having one defective fuse and two good fuses is 5%*95%*95%. Since the defective fuse could be any one of these three fuses, so we multiplied the aforementioned probability by 3 to indicate the three senarios (or you can say three orders).
Hopefully this helps XD