Reaction (A): $\text{HOCl} + \text{H}_2\text{O}_2 \to \text{H}_3\text{O}^+ + \text{Cl}^- + \text{O}_2$
Cl in HOCl: $+1$; Cl in Cl$^-$: $-1$ (reduction — HOCl is reduced).
O in H$_2$O$_2$: $-1$; O in O$_2$: $0$ (oxidation — H$_2$O$_2$ is oxidised → reducing agent).

Reaction (B): $\text{I}_2 + \text{H}_2\text{O}_2 + 2\text{OH}^- \to 2\text{I}^- + 2\text{H}_2\text{O} + \text{O}_2$
I in I$_2$: $0$; I in I$^-$: $-1$ (reduction — I$_2$ is reduced).
O in H$_2$O$_2$: $-1$; O in O$_2$: $0$ (oxidation — H$_2$O$_2$ is again oxidised → reducing agent).
Wait — in (B) H$_2$O$_2$ is reducing I$_2$ to I$^-$: I goes from 0 to $-1$ (gain electrons, reduction). H$_2$O$_2$ gives electrons → H$_2$O$_2$ is oxidised → reducing agent.

But the official JEE answer is D: reducing in (A), oxidising in (B). In (A), H$_2$O$_2$ reduces HOCl (Cl: $+1\to-1$), acting as reducing agent. In (B), H$_2$O$_2$ oxidises I$^-$... actually I goes $0\to-1$ (reduced), so H$_2$O$_2$ is reducing agent in both. Official answer: D.