Explanation

(a)
$\left\lbrack H^{+} \right\rbrack$for the solution of pH 3$= 1 \times 10^{- 3}$
$\left\lbrack H^{+} \right\rbrack$for the solution of pH 5$= 1 \times 10^{- 5}$
Let $V$ volumes of both the solutions are added, then concentration of $H^{+}$ in final mixture
$= \frac{1 \times 10^{- 3} \times V + 1 \times 10^{- 5} \times V}{V + V}$
$= \frac{V(1 \times 10^{- 3} + 1 \times 10^{- 5})}{2V}$
$= \frac{1.01 \times 10^{- 3}}{2}$
$= 5.05 \times 10^{- 4}$
pH of resultant solution $= - \log{\lbrack H^{+}\rbrack}$
$= - log(5.05 \times 10^{- 4})$
$= 3.29$
$\approx 3.3$