Answer: A, B, C
Acid hydration involves a free cation and a hydride shift, giving tertiary 2-methylbutan-2-ol. Oxymercuration is Markovnikov without rearrangement, giving 3-methylbutan-2-ol. Hydroboration is anti-Markovnikov, giving 3-methylbutan-1-ol. Hydroboration cannot give the tertiary alcohol.