Yes, that is what it is asking. I'm not surprised the proof was omitted since it is immediate from the expression you wrote down; you are just having trouble seeing the step (it happens to the best of us):

So we are just using the definition of conditional probability - the conditioning on X_n doesn't change anything in that regard.

Yes, this is true. A simple way to get at something like this would be2) For a discrete time Markov Chain, is it always true that

P(X_{n+2}=j | X_n =i ∩ X_{n-1}=s) = P(X_{n+2}=j | X_n =i) ??

I know that Markov's property implies that the above equality would always be true if the part in red is replaced by 1. But in this case, when we have 2 (instead of 1), would the equality still be true? If so, how can we formally prove it?

Thanks in advance!

then fiddle around with the rules of conditional probability to get rid of the conditioning on X_0.