How to change pH of CuSO4?
Alas, your problem has no solution, because of the risk of producing $\ce{Cu(OH)2}$ at pH $5$ and $6$. The solubility product of $\ce{Cu(OH)2}$ is $2.2·10^{-20}$. At pH $5$ and $6$, the concentrations of the ion $\ce{OH-}$ are respectively $10^{-9}$ and $10^{-8}$ M. This means that, at pH $5$ and $6$, the maximum molar concentrations of the ion $\ce{Cu^{2+}}$ are $2.2·10^{-2}$ M and $2.2·10^{-4}$ M respectively. So you will never be able to obtain a $0.1$ M solution of $\ce{CuSO4}$ at pH $5$ or $6$.