a)
$2K + 2H_2O \to 2KOH + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$
b)
Gọi $n_K = a(mol) ; n_{Na} = b(mol) \Rightarrow 39a + 23b = 8,5(1)$
Theo PTHH :
$n_{H_2} = 0,5a + 0,5b = \dfrac{3,36}{22,4} = 0,15(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2
$C_{M_{KOH}} = \dfrac{0,1}{0,2} = 0,5M$
$C_{M_{NaOH}} = \dfrac{0,2}{0,2} = 1M$