(a) Moment of inertia of sphere about any diameter $= \frac{2}{5} M R^{2}$ Using parallel axes theorem, Moment of inertia of sphere about a tangent to the sphere $= \frac{2}{5} M R^{2}+M R^{2}= \frac{7}{5} M R^{2}$ (b) We are given, moment of inertia of the disc about any of its diameter $= \frac{1}{4} M R^{2}$ (i) Applying perpendicular axes theorem, moment of inertia of the disc about an axis passing through its centre and normal to the disc $=2 \times \frac{1}{4} M R^{2}= \frac{1}{2} M R^{2}$ (ii) Applying parallel axes theorem, moment of inertia of the disc passing through a point on its edge and normal to the disc $= \frac{1}{2} M R^{2}+M R^{2}= \frac{3}{2} M R^{2}$