First problem (can’t figure out spoiler tags on my phone so my apologies):
If the snare hits 0 or 2 IHW, runner survives: 0 hit means their hand is left with 2 IHW only. First DRT hits both, leaving 6 cards, which is enough to survive the remaining 2 DRT. 2 hits mean the runner is left with 8 total cards after snare, enough to survive 3 DRT hits.
If snare hits 1 IHW, runner is left with 4 junk cards + 1 IHW. In order to die, IHW must be the last card remaining after the first two DRT hit.
Computing the above joint probability P(Snare hits 1 IHW ^ first two DRT miss IHW) = ((3 choose 2)*(2 choose 1)/(5 choose 3)) * 20% = 12%
Edit: can’t figure out spoiler tags while phone posting
Edit2: clarification: When snare hits, total possibilities are the binomial coefficient (5 choose 3). Outcomes where we hit one IHW can be counted by picking 2 cards out of 3, then 1 card out of 2. Thus we hit 1 IHW with probability (3 choose 2)*(2 choose 1)/(5 choose 3).