Let the downward direction be positive. The block moves a distance $x_m$ from initial contact with the spring to the point of maximum compression.

[Q1] The gravitational force $F_g = mg$ is constant and in the direction of displacement.

$$W_g = F_g d = mgx_m$$

[Q2] The spring force is directed opposite to the displacement. The work done by the spring is the negative of the change in its potential energy.

$$W_s = - \Delta U_s = - \left( \frac{1}{2}kx_m^2 - 0 \right) = -\frac{1}{2}kx_m^2$$

[Q3] We apply the work-energy theorem to the block during the compression phase. The net work done on the block equals its change in kinetic energy, $\Delta K = K_f - K_i$. The initial state is at impact ($v=v_i$) and the final state is at maximum compression ($v=0$).

$$W_{net} = W_g + W_s = \Delta K$$ $$mgx_m - \frac{1}{2}kx_m^2 = 0 - \frac{1}{2}mv_i^2$$

Solving for $v_i^2$:

$$v_i^2 = \frac{k}{m}x_m^2 - 2gx_m$$ $$v_i = \sqrt{\frac{kx_m^2}{m} - 2gx_m}$$

[Q4] For a new impact speed $v'_i = 2v_i$ and new compression $x'_m$, we use conservation of energy. Let the gravitational potential energy be zero at the initial contact point. Initial energy at impact: $E'_i = \frac{1}{2}m(v'_i)^2 = \frac{1}{2}m(2v_i)^2 = 2mv_i^2$. Final energy at new max compression: $E'_f = U_{g,f} + U_{s,f} = mg(-x'_m) + \frac{1}{2}k(x'_m)^2$. Setting $E'_i = E'_f$ and substituting the expression for $v_i^2$ from [Q3]:

$$2m\left(\frac{kx_m^2}{m} - 2gx_m\right) = -mgx'_m + \frac{1}{2}k(x'_m)^2$$ $$2kx_m^2 - 4mgx_m = -mgx'_m + \frac{1}{2}k(x'_m)^2$$

This is a quadratic equation for $x'_m$:

$$\frac{1}{2}k(x'_m)^2 - mg(x'_m) - (2kx_m^2 - 4mgx_m) = 0$$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ and taking the positive root for compression:

$$x'_m = \frac{mg + \sqrt{(-mg)^2 - 4(\frac{1}{2}k)(-(2kx_m^2 - 4mgx_m))}}{2(\frac{1}{2}k)}$$ $$x'_m = \frac{mg + \sqrt{m^2g^2 + 2k(2kx_m^2 - 4mgx_m)}}{k}$$ $$x'_m = \frac{mg + \sqrt{m^2g^2 + 4k^2x_m^2 - 8kmgx_m}}{k}$$