\(\begin{array}{l}
a)\\
Fe+CuSO_4\to FeSO_4+Cu\\
b)\\
n_{Fe}=\frac{1,96}{56}=0,035(mol)\\
n_{CuSO_4}=\frac{100.1,12.10\%}{160}=0,07(mol)\\
n_{Fe}<n_{CuSO_4}\to CuSO_4\,du\\
n_{FeSO_4}=0,035(mol);n_{CuSO_4(du)}=0,07-0,035=0,035(mol)\\
C_{M\,FeSO_4}=C_{M\,CuSO_4(du)}=\frac{0,035}{0,1}=0,35M\\
c)\\
n_{Cu}=n_{Fe}=0,035(mol)\\
m_{dd\,spu}=1,96+100.1,12-0,035.64=111,72(g)\\
C\%_{FeSO_4}=\frac{0,035.152}{111,72}.100\%\approx 4,76\%\\
C\%_{CuSO_4(du)}=\frac{0,035.160}{111,72}.100\%\approx 5,01\%
\end{array}\)