a,
PT điện li: $HCl\to H^{+}+Cl^-$
$\to [H^+]=C_{M_{HCl}}=0,02\ (M)$
$\to pH=-\log[H^+]=-\log(0,02)=1,7$
b,
PT điện li: $Ba(OH)_2\to Ba^{2+}+2OH^{-}$
$\to [OH^-]=2C_{M_{Ba(OH)_2}}=2.0,1=0,2\ (M)$
$\to [H^+]=\dfrac{10^{-14}}{[OH^-]}=\dfrac{10^{-14}}{0,2}=5.10^{-14}$
$\to pH=-\log[H^+]=-\log(5.10^{-14})=13,3$