I) Assuming that they are independent events you just apply the independence rule using the given probabilities.
P(E & T & J) = P(E) * P(T) * P(J) = 0.2 * 0.3 * 0.5 = 0.03
II) What you have to do is figure out the different combinations of hits and misses that results in only one of them hitting the bull's eye.
So we have: 1) Ev hits, Tracy misses and John misses. 2) Ev misses, Tracy hits and John misses. 3) Ev misses, Tracy misses and John hits.
Then you calculate the probability of each of those combinations happening using the probabilities you've been given. and you add them together. So for 1) It's 0.2 * 0.7 * 0.5 = 0.07. 2) It's 0.8 * 0.3 * 0.5 = 0.12. 3) It's 0.8 * 0.7 * 0.5 = 0.28. Then, finally, 0.07 + 0.12 + 0.28 = 0.47.
III) I'm pretty sure that you do the same thing you did in (II) but you find the combinations that involve one person missing and the combinations that involve no one missing. Then you add them together.
(II) and (III) are easy if you draw a probability tree and follow the branches along to find the probability of each combination and add them at the end.