a/ $Na+2H_2O\to 2NaOH+H_2$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
Theo phương trình: $n_{Na}=n_{H_2}$
$→n_{Na}=0,2(mol)\\→m_{Na}=0,2.23=4,6g$
Vậy $m_{Na}=4,6g$
b/ $m_{dd\,NaOH}=500.1,2=600g$
Theo phương trình: $n_{NaOH}=2n_{H_2}$
$→n_{NaOH}=0,4(mol)\\→m_{NaOH}=0,4.40=16g$
$\%C_A=\dfrac{16}{600}=2,66\%$
$C_{M_A}=\dfrac{0,4}{0,5}=0,8M$
Vậy $\%C_A=2,66\%,\,C_{M_A}=0,8M$