By theorem of parallel axes,
$$\eqalign{ & I = {I_{cm}} + M{d^2} \cr & I = {I_0} + M{\left( {\frac{L}{2}} \right)^2} = {I_0} + \frac{{M{L^2}}}{4} \cr} $$

The angular momentum of the mass $$m$$  about $$O$$  is $$m{r^2}\omega $$  and is directed toward $$+z$$  direction for all locations of $$m.$$

The angular momentum of mass $$m$$  about $$P$$  is $$mvl$$   and is directed for the given location of $$m$$  as shown in the figure.
The direction of $${{\vec L}_P}$$  remains changing for different locations of $$m.$$

The disc may be assumed as combination of two semicircular parts.
Let $$I$$  be the moment of inertia of the uniform semicircular disc
$$ \Rightarrow 2I = \frac{{2M{r^2}}}{2}\,\,\, \Rightarrow I = \frac{{M{r^2}}}{2}$$

Angular momentum of particle is given by :
$$L = m{r^2}\omega = 2\pi m{r^2}f\,\,\left[ {\because W = 2\pi f} \right]$$
If frequency is halved then,
$$L' = m{r^2}\frac{\omega }{2} = \pi m{r^2}f\,\,\therefore L' = \frac{L}{2}$$

Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane is $${I_{AB}} = \frac{1}{2}M{R^2}$$

Using theorem of parallel axes, we have,
$$\eqalign{ & {I_{CD}} = {I_{AB}} + M{R^2} \cr & = \frac{1}{2}M{R^2} + M{R^2} \cr & = \frac{3}{2}M{R^2} \cr} $$

Radius of gyration of a body depends on the axis of rotation.

Since, in this case, instantaneous axis of rotation is always below the centre of mass. This is possible only when point of contact moves with a velocity equal to centre of mass.

$$\eqalign{ & {\text{For solid sphere}} \cr & {I_{AB}} = \frac{2}{5}M{R^2} = I\,\,\,\left( {{\text{Given}}} \right).....(i) \cr & {\text{For solid dice}} \cr & {I_{A'B'}} = {I_{YY'}} + M{r^2} = \frac{1}{2}M{r^2} + M{r^2} = \frac{3}{2}M{r^2} \cr & {I_{AB}} = {I_{A'B'}}\,\,\,\,\,\left( {{\text{Given}}} \right).....(ii) \cr & {\text{From }}(i)\,\,{\text{and }}(ii), \cr & \frac{2}{5}M{R^2} = \frac{3}{2}M{r^2} \cr & \Rightarrow r = \frac{2}{{\sqrt {15} }}R \cr} $$

When two small spheres of mass $$m$$ are attached gently, the external torque, about the axis of rotation, is zero and therefore the angular momentum about the axis of rotation is constant.
$$\eqalign{ & \therefore {I_1}{\omega _1} = {I_2}{\omega _2} \Rightarrow {\omega _2} = \frac{{{I_1}}}{{{I_2}}}{\omega _1} \cr & {\text{Here }}{I_1} = \frac{1}{2}M{R^2}{\text{ and }}{I_2} = \frac{1}{2}M{R^2} + 2m{R^2} \cr & \therefore {\omega _2} = \frac{{\frac{1}{2}M{R^2}}}{{\frac{1}{2}M{R^2} + 2m{R^2}}} \times {\omega _1} = \frac{M}{{M + 4m}}{\omega _1} \cr} $$

As initially both the particles were at rest therefore velocity of centre of mass was zero and there is no external force on the system so speed of centre of mass remains constant i.e it should be equal to zero.