As noted in the comments to your question, Stack Exchange frowns on answering these type of questions that lack any demonstration of work so far as they read like homework questions (or interview questions in this case) on which you've made no effort to solve. But they're fun, so typical stack exchange chiding aside here's my try at what are three interesting problems:
1) The answer is roughly 4, as estimated by simulation in Python:
import pandas as pd
i = 1
j = 1
i = max(0, i + random.sample([-1,1], 1))
j += 1
if i == 2:
i = 1
j += 1
ants = 
i = 0
while i < 100000:
i += 1
df = pd.DataFrame(ants, columns = ['turns'])
print(df.mean()) #About 4
The key here is that if the ant can walk closer (+1) or farther (-1) away, but is never gets more than 2 moves from its destination and starts 2 moves from its destination. You also need to factor in that if two moves away the ant always moves closer (because they're moving around a square rather than along a line). So the
else statement adds this handicap by tacking on a move and starting them at 1 if they end up 0 (which represents their original starting point from which they can only get closer to the destination).
For 2, I believe you copied the problem incorrectly (it typically asks about the optimal strategy for the fewest number of drops). As you ask the question, you only need one egg. You just precede from 1 to 2 to 3, to 4, and when the egg breaks at n you know the threshold floor. This is assuming a perfectly deterministic rather than stochastic process, i.e. the egg either always or never breaks at a given floor, which is probably not true in the real world. If you account for randomness, then you need to consider the possibility that an egg will crack below the floor that it's guaranteed to crack at with probability 1. So I'll let someone else take a stab at how to estimate the threshold floor with two eggs and provide more practical advice: if you don't want eggs to break stop dropping them out of a window from any story.
3) If you spin you have a straightforwand 2/6 = 1/3 chance of biting a bullet. If you do not spin there are two bullets and five slots, except that you know you're not about to bite bullet 2 (in order) because you did not just bite bullet 1, so you can rule bullet 2 out and you thus have a 1/4 chance of getting shot if he does not spin. So don't spin.