\(a.n_{Al.pư}=a\\ 2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\\ \Rightarrow m_{\Delta}=\dfrac{3}{2}\cdot64a-27a=46,38-45\\ \rightarrow a=0,02mol\\ n_{Cu}=n_{CuSO_4}=\dfrac{3}{2}n_{Al}=0,03mol\\ m_{Cu}=0,03.64=1,92g\\ b.C_{M_{CuSO_4}}=\dfrac{0,03}{0,4}=0,075M\)