Is this a correct way to show that a problem is coNP-complete? Let $A$ be a problem that I want to show it is coNP-complete.
I know I could just show its  complement $\bar{A}$ is NP-complete
 or that $\bar{A}$ is in NP and for some coNP-complete problem $Q$, show that $Q$ can be Karp-reduced to $A$.

But I wonder if the following steps are sufficient to show that A is coNP-complete? 


   Show that $A$ is in $coNP$ by showing that its complement is in NP
   Choose one coNP-complete problem $Q$ and Cook-reduce it to $A$. 


Does the coNP-complete class still be distinguishable (under the assumption $P\neq NP$) from the NP-complete class? since it seems that any coNP-complete problem is Turing-polynomially equivalent to any other NP-Complete problem.
No, the steps are insufficient. You need to use a Karp reduction, since that's the type of reduction used in the definition of coNP-completeness. 

SAT is NP-complete, but probably not coNP-complete (unless NP=coNP); and coSAT is coNP-complete, but probably not NP-complete. This distinction would be lost if you used Cook reductions. 

Some questions on this site which are relevant are this one and that one, as well as a few others.