$a) 2NaOH + CuSO_4 \to Cu(OH)_2 + Na_2SO_4$
$b) n_{NaOH} = \dfrac{8}{40} = 0,2(mol)$
Theo PTHH, $n_{Cu(OH)_2} = \dfrac{1}{2}n_{NaOH} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$
$c) n_{CuSO_4} = \dfrac{1}{2}n_{NaOH} = 0,1(mol0$
$\Rightarrow C_{M_{CuSO_4}} = \dfrac{0,1}{0,04} = 2,5M$