This one doesn't have a prohibition on asking tricky questions.
If we assume that a lie is the opposite of what the truth telling guard would say, and that the number of safe doors is known, you can ask one guard one question: "Which doors are safe?". If he points at the correct number of doors, then you go through one that was pointed at, otherwise you go through a door not pointed at. Whether the merciless guard lies or not doesn't matter, because you know if the statement is truth or lie automatically.
However this version specifically allows asking multiple questions, and doesn't prohibit asking tricky questions.
A "tricky question" is a useful question that forces the liar to answer how the truth teller would. This includes any questions about future answers, or any trickery you would use to weasel out of it. But, "Are you the truth teller?" isn't a tricky question, even though the liar would give the same answer, because it doesn't actually help at all.
Therefore, if we do not know the number of safe doors, we can ask the following tricky question of two guards.
"If I were to ask you which doors are safe, which ones would you point at?" Both the truth teller and the liar will give the same answer, while the merciless guard can give either answer. If the two answers are the same, they are correct, and there is no need to ask the third guard. If they are different, then one of them is the merciless guard, and thus the third is either the truth teller or the liar, and thus will answer the question properly when asked it of them. Simple.
If we add back in the "no tricky questions" constraint from the comic, then the problem becomes trickier.
If the merciless guard is required to alternate between truth and lie, then the problem is solvable after all by the previously listed method. Or by asking a truth probing non tricky question of each guard twice to identify them, then asking the truth teller.
If not, there's still a solution. You first ask each guard how many fingers you are holding up. If the merciless guard lies, you've identified the truth teller, and can ask him what doors are safe. If the merciless guard tells the truth, you've identified the liar. you then ask the liar if another guard is the merciless guard. he will say no if it is, and yes if it isn't. You've now identified the truth teller, and only asked one question of the merciless guard.
Now if a liar can say they don't know when they do, then you are pretty screwed, because you can't do this trick. The liar will simply refuse to help you figure out who the merciless guard is, and the merciless guard can lie or tell the truth as needed to screw you over.
But in this case, the limit of two questions to the merciless guard isn't even needed. It doesn't matter how may questions you ask, you can't solve it without a tricky question.
Therefore we can presume that the guard is not free to make its own decision about whether to lie or tell the truth. If it can repeat an answer to the same question, you can't solve it because of the 2 question limit. But if it can't, then you only need to ask it the same question twice to identify it, and not ask it a third question.