Explanation

(d)
Moles of $HCl = \frac{0.365}{36.5} = 0.01$
Moles of $NaOH\ in\ 100\ cm^{3}\ of\ 0.2\ M\ NaOH$
$0.2 = \frac{n \times 1000}{100} = 0.02$
Moles of $NaOH$ left=0.02$-$0.01=0.01
Moles of $NaOH$ per litre
$\frac{0.01 \times 1000}{100} = 0.1$
$\left\lbrack H^{+} \right\rbrack = \frac{10^{- 14}}{0.1} = 10^{- 13};\ \ (K_{w} = \left\lbrack H^{+} \right\rbrack.\left\lbrack OH^{-} \right\rbrack)$
$pH = - \log{\left\lbrack H^{+} \right\rbrack = - \log\left\lbrack 10^{- 13} \right\rbrack}$
$pH = 13$