The following diagram shows the solution to the mechanistic question. Note that all the information applies to a single reaction sequence that has been completely described verbally.  There is no need for extra reagents or extra steps etc. The curly arrows are drawn specifically to match the text in the question. The biggest problem students have is making sure they understand the language of chemistry. Most students have trouble because they can't draw the structures from the IUPAC names (that means they don't know their nomenclature well enough). Read the words carefully, and then make the curly arrows tell that same story. There is NO need for extra steps. Remember curly arrows go from electron rich to poor and to balance the formal charges at each step - errors on formal charges were common.

a.

b. The alkene reacts with the acid to give the same tertiary carbocation. But since there is no water present, a different nucleophile, the chloride ion, reacts to give 1-chloro-1-methylcyclohexane. This part of the question is about recognising the nucleophile.

c. Why is aqueous hydrochloric acid a poor choice? If you look at the answers to parts a and b, you see that the difference is due to the nucleophile. Hence aq. HCl could form two products, the alcohol from a and the alkyl chloride from b since both water and chloride would be present. Note that under aqueous conditions, it is H3O+ that is the acid present for either aq. H2SO4 or aq. HCl.

d. Compare to the reaction in part a which states that it gives the more stable carbocation. Look at the 1-phenylpropene and consider the two possible carbocations...by applying the principles of resonance, we can predict that the one carbocation is benzylic. Hence the major product is 1-phenyl-1-propanol. From the hydrocarbon laboratory experiment, you should remember that aromatic rings are less reactive than the alkene so it's the alkene that reacts.