a) $n_{NaOH} = \dfrac{8}{40} = 0,2(mol)$
$n_{OH^-} = n_{NaOH} = 0,2(mol)$
$[OH^-] = \dfrac{0,2}{2} = 0,1M$
$[H^+] = \dfrac{10^{-14}}{[OH^-]} = 10^{-13}M$
b)
$n_{HCl} = \dfrac{1,825}{36,5} = 0,05(mol)$
$n_{H^+} = n_{HCl} = 0,05(mol)$
$[H^+] = \dfrac{0,05}{0,5} = 0,1M$
$[OH^-] = \dfrac{10^{-14}}{[H^+]} = \dfrac{10^{-14}}{0,1} = 10^{-13}(M)$