There are a few parts to this question, so let's break it down.
First the problem tells you that CuSO4*5H2O is reacted with NaOH until all the copper sulfate is precipitated into Cu(OH)2. This means that the copper sulfate must be the limiting reagent, and if we find the moles of copper sulfate, then that is the moles of Cu(OH)2. So, to find the number of moles from the information given (12.07 g), we need to figure out the molar mass.
Cu = 63.55, S = 32.06, 4O = 4(16.00), 10H = 10(1.01), 5O = 5(16.00). The total molar mass is 249.71 g/mol.
Now to find the number of moles, divide 12.07 g by 249.71 g/mol. This will cancel out grams and leave moles on top, which is the unit we want. 12.07/249.71 = 0.04834 mol. Since CuSO4 and Cu(OH)2 must be in a one to one ratio, that value we calculated is also the number of moles of Cu(OH)2.
Now, the problem says that 35.27% of Cu(OH)2 was converted to CuO, which means that 64.73% is left behind or is left unreacted. To get the amount of moles left behind, you need to multiply the decimal with the number of starting moles. 0.04834*0.6473 = 0.03129 mol. Presumably, the question will ask for the answer in grams, so all you need to do now is multiply the molar mass of Cu(OH)2 by 0.03129 mol. This will cancel out moles and leave grams on top.