Hardness and approximation of a problem with a parameter
Yes.  To show that decision problem Q is at least as hard as decision problem P (via a many-one reduction), you need to prove that a yes-instance for P maps to a yes-instance for Q, and a no-instance for P maps to a no-instance for Q.  That's just what the definition of a (many-one) reduction is.

In your case, P = "X has a solution with parameter $k'$" and Q = "H has a solution with parameter $k$".  Therefore, the answer to your question is yes: in (1) you need to prove that $X$ has no solution with parameter $k'$.  The definition of P needs to be consistent for both kinds of instances, so it has to use the same parameter in both cases.