The system is a physical pendulum. The distance from the pivot to the center of mass (the disk's center) is $d=L+R=L+D/2$. By the parallel axis theorem, the moment of inertia about the pivot is $I = I_{cm} + md^2 = \frac{1}{2}mR^2 + m(L+R)^2$.

[Q1] Without the spring, the period $T_p$ is:

$$T_p = 2\pi \sqrt{\frac{I}{mgd}} = 2\pi \sqrt{\frac{\frac{1}{2}mR^2 + m(L+R)^2}{mg(L+R)}} = 2\pi \sqrt{\frac{R^2/2 + (L+R)^2}{g(L+R)}}$$

[Q2] With the spring, the total restoring torque for a small angle $\theta$ is $\tau = -(mgd)\theta - \kappa\theta = -(mgd + \kappa)\theta$. The equation of motion is $I\ddot{\theta} = \tau$. The new period $T_{new}$ is:

$$T_{new} = 2\pi \sqrt{\frac{I}{mgd + \kappa}}$$

Given $T_{new} = T_p - \Delta t$, where $\Delta t = 0.500$ s, we solve for the torsion constant $\kappa$:

$$mgd + \kappa = \frac{4\pi^2 I}{T_{new}^2} \implies \kappa = \frac{4\pi^2 I}{(T_p - \Delta t)^2} - mgd$$