(N/A) Part $(a)$: According to the law of conservation of angular momentum,the total angular momentum of the system remains constant as no external torque acts on the system.
Initial angular momentum $L_{i} = I_{1}\omega_{1} + I_{2}\omega_{2}$.
When the discs are joined,they rotate with a common angular speed $\omega$. The total moment of inertia of the system becomes $I = I_{1} + I_{2}$.
Final angular momentum $L_{f} = (I_{1} + I_{2})\omega$.
Equating $L_{i} = L_{f}$,we get: $I_{1}\omega_{1} + I_{2}\omega_{2} = (I_{1} + I_{2})\omega$.
Therefore,the angular speed of the system is $\omega = \frac{I_{1}\omega_{1} + I_{2}\omega_{2}}{I_{1} + I_{2}}$.
Part $(b)$: Initial kinetic energy $E_{i} = \frac{1}{2}I_{1}\omega_{1}^{2} + \frac{1}{2}I_{2}\omega_{2}^{2}$.
Final kinetic energy $E_{f} = \frac{1}{2}(I_{1} + I_{2})\omega^{2} = \frac{1}{2}(I_{1} + I_{2})\left(\frac{I_{1}\omega_{1} + I_{2}\omega_{2}}{I_{1} + I_{2}}\right)^{2} = \frac{(I_{1}\omega_{1} + I_{2}\omega_{2})^{2}}{2(I_{1} + I_{2})}$.
The loss in kinetic energy is $\Delta E = E_{i} - E_{f} = \frac{1}{2}I_{1}\omega_{1}^{2} + \frac{1}{2}I_{2}\omega_{2}^{2} - \frac{(I_{1}\omega_{1} + I_{2}\omega_{2})^{2}}{2(I_{1} + I_{2})}$.
Simplifying this expression,we get $\Delta E = \frac{I_{1}I_{2}(\omega_{1} - \omega_{2})^{2}}{2(I_{1} + I_{2})}$.
Since $I_{1}, I_{2} > 0$ and $(\omega_{1} - \omega_{2})^{2} > 0$ (as $\omega_{1} \neq \omega_{2}$),$\Delta E > 0$,which implies $E_{i} > E_{f}$.
This loss in kinetic energy is due to the work done against the frictional force that acts between the surfaces of the discs until they attain a common angular velocity.