If a problem is Cook-NP hard, and this problem is in NP, does it prove that the problem is Karp-NP-complete?
NP-hardness is a property of decision problems. It doesn't matter whether you define it using Karp reductions or Cook reductions.

NP-hardness and NP-completeness are defined using Karp reductions. If you prove that a problem is Cook-NP-hard, then it doesn't imply that it is Karp-NP-hard. Indeed, coSAT (the collection of unsatisfiable CNFs) is Cook-NP-hard but probably not Karp-NP-hard. Under reasonable hardness assumptions, there are even problems in NP which are Cook-NP-hard but not Karp-NP-hard. See for example Mandal, Pavan, and Venugopalan, Separating Cook Completeness from Karp-Levin
Completeness under a Worst-Case Hardness
Hypothesis.