$$\eqalign{ & {\text{Using}}\,{v^2} = {\omega ^2}\left( {{a^2} - {y^2}} \right)\,{\text{we}}\,{\text{have}} \cr & {10^2} = {\omega ^2}\left( {{a^2} - {4^2}} \right) \cr & {\text{and}}\,\,{8^2} = {\omega ^2}\left( {{a^2} - {5^2}} \right); \cr & {\text{so}}\,\,{10^2} - {8^2} = {\omega ^2}\left( {{5^2} - {4^2}} \right) = \left( {3{\omega ^2}} \right)\,{\text{or}}\,6 = 3\,\omega \cr & {\text{or}}\,\omega = 2\,{\text{or}}\,T = \frac{{2\,\pi }}{\omega } = \frac{{2\,\pi }}{2} = \pi \,s. \cr} $$

The equation of motion for the pendulum, suffering retardation
$$\eqalign{ & I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right){\text{where}}\,I = m{\ell ^2}\,{\text{and}}\,\alpha = \frac{{{d^2}\theta }}{{d{t^2}}} \cr & \therefore \frac{{{d^2}\theta }}{{d{t^2}}} = \frac{g}{\ell }\tan \theta + \frac{{bv}}{\ell } \cr} $$
On solving we get $$\theta = {\theta _0}{e^{ - \frac{{bt}}{2}\sin \left( {\omega t + \phi } \right)}}$$
According to questions $$\frac{{{\theta _0}}}{e} = {\theta _0}{e^{ - \frac{{b\tau }}{2}}}$$
$$\therefore \tau = \frac{2}{b}$$

$$\eqalign{ & {\text{As we know,}}\,E = {E_0}{e^{ - \frac{{bt}}{m}}} \cr & 15 = 45{e^{ - \frac{{b15}}{m}}} \Rightarrow \frac{1}{3} = {e^{ - \frac{{b15}}{m}}}\,\,\left[ {{\text{As no}}{\text{. of oscillations}} = 15{\text{ so}}\,t = 15\,\sec } \right] \cr} $$
Taking log on both sides
$$\frac{b}{m} = \frac{1}{{15}}\ln 3$$

Maximum velocity,
$$\eqalign{ & {v_{\max }} = a\omega ,\,{v_{\max }} = a \times \frac{{2\pi }}{T} \cr & \Rightarrow T = \frac{{2\pi a}}{{{v_{\max }}}} = \frac{{2 \times 3.14 \times 7 \times {{10}^{ - 3}}}}{{4.4}} \approx 0.01\,s \cr} $$

For a particle executing $$SHM,$$  we have maximum acceleration,
$$\alpha = A{\omega ^2}\,......\left( {\text{i}} \right)$$
where, $$A$$ is maximum amplitude and $$\omega $$ is angular velocity of a particle.
Maximum velocity, $$\beta = A\omega \,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (i) by Eq. (ii), we get
$$\eqalign{ & \frac{\alpha }{\beta } = \frac{{A{\omega ^2}}}{{A\omega }} \cr & \Rightarrow \frac{\alpha }{\beta } = \omega = \frac{{2\pi }}{T} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,T = \frac{{2\pi \beta }}{\alpha } \cr} $$
Thus, its time period of vibration, $$T = \frac{{2\pi \beta }}{\alpha }$$

$${E_{av}} = {U_{av}} = \frac{1}{4}m{\omega ^2}{A^2}$$

$$\eqalign{ & x = A\sin \left( {\omega t + \phi } \right) \cr & {\text{where}}\,\,A = \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 - {\omega ^2}} \right)}^2}} }} = \frac{{{F_0}}}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}} \cr} $$
Here damping effect is considered to be zero
$$\therefore x \propto \frac{1}{{m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$

Expression of kinetic energy is
$$K = \frac{1}{2}k\left( {{a^2} - {x^2}} \right)\,......\left( {\text{i}} \right)$$
Expression of potential energy is
$$U = \frac{1}{2}k{x^2}\,......\left( {{\text{ii}}} \right)$$
where, $$k = m{\omega ^2}$$
We observe that at mean position $$\left( {x = 0} \right),$$   kinetic energy is maximum $$\left( {\frac{1}{2}k{a^2}} \right)$$  and potential energy is minimum {zero}. Also at extreme positions $$\left( {x = \pm a} \right),$$   kinetic energy is zero and potential energy is maximum $$\left( {\frac{1}{2}k{a^2}} \right).$$  Thus, displacement between positions of maximum potential maximum kinetic energy is $${ \pm a}.$$
NOTE
Kinetic energy is zero at extreme positions but potential energy at mean position need not be zero. It is minimum at mean position.

Potential energy is given by $$U = \frac{1}{2}k{x^2}$$
The corresponding graph is shown in figure.

At equilibrium position $$\left( {x = 0} \right),$$   potential energy is minimum. At extreme positions $${x_1}$$ and $${x_2},$$ its potential energies are
$${U_1} = \frac{1}{2}kx_1^2\,\,{\text{and}}\,\,{U_2} = \frac{1}{2}kx_2^2$$
NOTE
In the above graph, the dotted line (curve) is shown for kinetic energy. This graph shows that kinetic energy is maximum at mean position and zero at extreme positions $${x_1}$$ and $${x_2}.$$

At point 2, the acceleration of the particle is maximum, which is at the extreme position. At extreme position, the velocity of the particle will be zero.