To determine the role of $H_2O_2$ in each reaction, we need to analyze the changes in oxidation states.$\\$(a) Reaction: $H_2O_2 + O_3 \rightarrow H_2O + 2O_2 \\$• In $H_2O_2,$ the oxidation state of oxygen is $-1. \\$• In $O_3,$ the oxidation state of oxygen is $0. \\$• In $H_2O,$ the oxidation state of oxygen is $-2. \\$• In $O_2,$ the oxidation state of oxygen is $0. \\$In this reaction, $H_2O_2$ is converted to $H_2O$ and $O_2. \\$The oxygen in $H_2O_2$ goes from $-1$ to $-2$ in $H_2O$ (reduction) and $-1$ to $0$ in $O_2$ (oxidation).$\\$Since $H_2O_2$ is being reduced, it acts as a reducing agent.$\\$(b) Reaction: $H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2 \\$• In $Ag_2O,$ the oxidation state of silver is $+1. \\$• In $Ag,$ the oxidation state of silver is $0. \\$• In $H_2O_2,$ the oxidation state of oxygen is $-1. \\$• In $O_2,$ the oxidation state of oxygen is $0. \\$In this reaction, $Ag_2O$ is reduced to $Ag$ and $H_2O_2$ is oxidized to $O_2. \\$Since $H_2O_2$ is being oxidized, it acts as a reducing agent.$\\$Therefore, in both reactions, $H_2O_2$ acts as a reducing agent.$\\$The correct option is:$\\$Option 3: Reducing in (a) and (b)