It is noticed that Pb^ 2+ is more stable than Pb^ 4+ but Sn^ 2+ is...

The stability of oxidation states in Group 14 is governed by the inert pair effect, which increases down the group.
For Lead ($Pb$), the $+2$ oxidation state is more stable than the $+4$ state ($Pb^{2+} > Pb^{4+}$).
In the reaction $PbO_2 + Pb \rightarrow 2PbO$, $Pb^{4+}$ is being reduced to $Pb^{2+}$ and $Pb^0$ is being oxidized to $Pb^{2+}$. Since the product contains the more stable oxidation state, the reaction is spontaneous, meaning $\Delta_r G^\circ(1) < 0$.
For Tin ($Sn$), the $+4$ oxidation state is more stable than the $+2$ state ($Sn^{4+} > Sn^{2+}$).
In the reaction $SnO_2 + Sn \rightarrow 2SnO$, the stable $Sn^{4+}$ is being converted into the less stable $Sn^{2+}$. This process is non-spontaneous, meaning $\Delta_r G^\circ(2) > 0$.
Thus, the correct set is $\Delta_r G^\circ(1) < 0$ and $\Delta_r G^\circ(2) > 0$.