\(1) m_{NaCl} =\dfrac{200.5}{100} = 10 (g)\\ \rightarrow m_{H_2O} = 200 - 10=190(g)\\ 2)m_{CuSO_4} = \dfrac{8.500}{100} = 40 (g)\\ \rightarrow n_{CuSO_4} = \dfrac{40}{160} = 0,25 (mol)\\ m_{H_2O} = 500 - 40 = 460(g)\\ 3) m_{NaOH} = 1,2.40 = 48 (g)\\ \rightarrow m_{ddNaOH} = \dfrac{48.100}{15} = 320 (g)\\ \rightarrow m = 320 - 48 = 272 (g)\)