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Under example 1, I'm not sure what you mean about accounting for all red boxes. In theory, a blue chest claiming there's more red mimics than blue mimics could still be a mimic if there's 1 red mimic, because then there would also be 1 blue mimic so it would be lying. The reason it can't be the third mimic in that example is because it's too far away from the truthful 1 in the east position.
Under example 3, the middle box doesn't just say there's a blue mimic, it specifically says there's 1 blue mimic. That plus the two blue chests in east and southeast both being safe means exactly one of the corner 1s is a mimic and the other is safe.
When you have two 0s in opposite corners in a 3-mimic puzzle, it's true that they can't both be telling the truth. There is one way they could both be mimics, and that's if the center chest is the third mimic. So if you've got two 0s in opposite corners and you've worked out that one of them is definitely a mimic and the center chest is definitely safe, the other 0 will also be safe.
In a 2-mimic puzzle, they can't both be mimics, and the only way they could both be truthful is if the other two corners are the mimics. So the vast majority of the time, once you've worked out what one of them is, you immediately know what the other one is too.
In this case, it was the second one.