This is a standard synthesis problem. It tells you the name of the starting material, that it takes 3 steps, and the end product and you are asked to fill in the intermediate products and reagents. I will go through it in words and then attach a screenshot of the problem below.
You are told the starting material is 2-methyl-1-butene and the end product will be 3-bromo-2-methyl-2-butanol. First thing I looked at was the positioning of the substituents on the final product. Based on this, I knew I was going to need a double bond between the bromine and hydroxyl group (carbons 2-3). To do this, the first reagent could be HBr and CH2Cl2, which would result in a tertiary bromine (2-bromo-2-methyl butane). Then if you add EtO-K+, a strong base, a double bond would be formed between carbons 2-3 via an E2 elimination reaction (2-methyl-2-butene). You know the double bond will form there because it results in the most substituted alkene. The next and final reagent needed for the product is Br2 and H2O. The OH will go to the more substituted carbon due to the more partial postive carbon intermediate. This results in the wanted final product: 3-bromo-2-methyl-2-butanol.