Let T be the tension in the cord and the downward direction be positive. Apply Newton's second law to the block:

$$F_{net} = Mg - T = Ma$$

Given the acceleration $a = g/4$:

$$Mg - T = M(g/4)$$ $$T = Mg - Mg/4 = \frac{3}{4}Mg$$

[Q1] The cord's force (tension) is directed upward, opposite to the downward displacement $d$. The work done by the cord, $W_T$, is:

$$W_T = -Td = -\frac{3}{4}Mgd$$

[Q2] The gravitational force $Mg$ is in the same direction as the displacement $d$. The work done by gravity, $W_g$, is:

$$W_g = Mgd$$

[Q3] The net work done on the block equals the change in its kinetic energy (Work-Energy Theorem), $W_{net} = \Delta K$. The block starts from rest, so its initial kinetic energy $K_i=0$.

$$K_f = W_{net} = W_g + W_T$$ $$K_f = Mgd - \frac{3}{4}Mgd = \frac{1}{4}Mgd$$

[Q4] The kinetic energy is also given by $K_f = \frac{1}{2}Mv^2$. We can solve for the speed $v$:

$$\frac{1}{2}Mv^2 = \frac{1}{4}Mgd$$ $$v^2 = \frac{gd}{2}$$ $$v = \sqrt{\frac{gd}{2}}$$