The correct option is D both (a) & (b)
As we move down the group, II oxidation state become more stable than IV so Pb II is more stable than Pb IV and PbO2 has high tendency to get electron and converts into PbO.
So as stability of oxidation states,
PbII>PbIV and SnIV>SnII.
So for first reaction, free energy is less than zero and for second reaction free energy is positive.