$Zn+CuSO_4\to ZnSO_4+Cu$
$Zn+CdSO_4\to ZnSO_4+Cd$
$n_{CuSO_4}=\dfrac{3,2}{160}=0,02(mol)$
$n_{CdSO_4}=\dfrac{6,4}{208}=0,03(mol)$
$\to \begin{cases} n_{Zn\text{pứ}}=0,02+0,03=0,05(mol)\\ n_{Cu}=0,02(mol)\\ n_{Cd}=0,03(mol)\end{cases}$
$\to m_{\text{thanh kẽm tăng}}=0,02.64+0,03.112-0,05.65=1,39g$
Vậy thanh kẽm tăng $1,39g$