$n_{Na_2O}=6,2/62=0,1mol$
$PTHH :$
$Na_2O+H_2O\to 2NaOH$
$\text{a.Theo pt : }$
$n_{NaOH}=2.n_{Na_2O}=2.0,1=0,2mol$
$⇒C_{M_{Na_2O}}=\dfrac{0,2}{0,1}=2M$
$b.2NaOH+H_2SO_4\to Na_2SO_4+2H_2O(*)$
$\text{Theo pt :}$
$n_{H_2SO_4}=1/2.n_{NaOH}=1/2.0,2=0,1mol$
$⇒V_{H_2SO_4}=\dfrac{0,1}{2}=0,05l$
$\text{c.Theo pt (*) :}$
$n_{Na_2SO_4}=1/2.n_{NaOH}=0,1mol$
$⇒n_{Na_2SO_4.10H_2O}=n_{Na_2SO_4}=0,1mol$
$⇒m_{Na_2SO_4.10H_2O}=0,1.322=32,2g$