(N/A) $Pb^{4+}$ by gaining $2$ electrons changes into $Pb^{2+}$,which is more stable due to the inert pair effect. Thus,it acts as an oxidizing agent.
$Sn^{2+}$ is less stable than $Sn^{4+}$ and tends to lose $2$ electrons to form $Sn^{4+}$,thus acting as a reducing agent.
$(1)$ $Pb^{4+} + 2e^{-} \rightarrow Pb^{2+}$ (Reduction,hence oxidizing agent)
$Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Oxidation,hence reducing agent)
$(2)$ The electron gain enthalpy of fluorine is less negative than that of chlorine because of the small size of the fluorine atom. The inter-electronic repulsions in the compact $2p$ subshell of fluorine make the addition of an electron less favorable compared to the $3p$ subshell of chlorine.