a,
$n_{H_2SO_4}=\dfrac{58,8.20\%}{98}=0,12(mol)$
$m_{AOH}=134,4.10\%=13,44g$
$2AOH+H_2SO_4\to A_2SO_4+2H_2O$
$\Rightarrow n_{AOH}=0,12.2=0,24(mol)$
$M_{AOH}=\dfrac{13,44}{0,24}=56=M_A+17$
$\Leftrightarrow M_A=39(K)$
$\to A$ là kali
b,
$n_{K_2SO_4}=0,12(mol)$
$\Rightarrow C_{M_{K_2SO_4}}=\dfrac{0,12}{0,4}=0,3M$