$$\eqalign{ & {\text{Max}}{\text{. force}} = {\text{mass}} \times {\text{max}}{\text{. acceleration}} \cr & = m\,4{\pi ^2}{v^2}a = 1 \times 4 \times {\pi ^2} \times {\left( {60} \right)^2} \times 0.02 \cr & = 288\,{\pi ^2} \cr} $$

Let us consider two springs of spring constants $${{k_1}}$$ and $${{k_2}}$$ joined in series as shown in figure.
Under a force $$F,$$ they will stretch by $${y_1}$$ and $${y_2}.$$
So, $$y = {y_1} + {y_2}$$

$${\text{or}}\,\,\frac{F}{k} = \frac{{{F_1}}}{{{k_1}}} + \frac{{{F_2}}}{{{k_2}}}$$
But as springs are massless, so force on them must be same, i.e. $${F_1} = {F_2} = F.$$
$$\eqalign{ & {\text{So,}}\,\,\frac{1}{k} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \cr & {\text{or}}\,\,k = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} \cr} $$

$$K.E.$$  of a body undergoing SHM is given by,
$$K.E. = \frac{1}{2}m{a^2}{\omega ^2}{\cos ^2}\omega t,\,T.E. = \frac{1}{2}m{a^2}{\omega ^2}$$
Given $$K.E. = 0.75\,T.E.$$
$$\eqalign{ & \Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = \frac{\pi }{6} \cr & \Rightarrow t = \frac{\pi }{{6 \times \omega }} \Rightarrow t = \frac{{\pi \times 2}}{{6 \times 2\pi }} \Rightarrow t = \frac{1}{6}s \cr} $$

$$\eqalign{ & {v_1} = \frac{{d{y_1}}}{{dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + \frac{\pi }{3}} \right) \cr & {v_2} = \frac{{d{y_2}}}{{dt}} = - 0.1\pi \sin \pi t = 0.1\pi \cos \left( {\pi t + \frac{\pi }{2}} \right) \cr & \therefore {\text{Phase diff}}{\text{.}} = {\phi _1} - {\phi _2} = \frac{\pi }{3} - \frac{\pi }{2} = \frac{{2\pi - 3\pi }}{6} = - \frac{\pi }{6} \cr} $$

In case (ii), the springs are shown in the maximum compressed position. If the spring of spring constant $${{k_1,}}$$  is compressed by $${{x_1}}$$ and that of spring constant $${{k_2}}$$ is compressed by $${{x_2}}$$ then
$$\eqalign{ & {x_1} + {x_2} = A\,......\,\left( {\text{i}} \right) \cr & {\text{and }}{k_1}{x_1} = {k_2}{x_2} \Rightarrow {x_2} = \frac{{{k_1}{x_1}}}{{{k_2}}}\,......\,\left( {{\text{ii}}} \right) \cr & {\text{From}}\left( {\text{i}} \right)\& \left( {{\text{ii}}} \right) \cr & {x_1} + \frac{{{k_1}{x_1}}}{{{k_2}}} = A \Rightarrow {x_1} = \frac{{{k_2}A}}{{{k_2} + {k_1}}} \cr} $$

$$\eqalign{ & {v_1} = \omega \sqrt {{a^2} - x_1^2} ,{v_2} = \omega \sqrt {{a^2} - x_2^2} \cr & {\text{We}}\,{\text{get,}}\,\,a = \sqrt {\frac{{v_1^2x_2^2 - v_2^2x_1^2}}{{v_1^2 - v_2^2}}} \cr} $$