Assertion (A): The work done by a force is (W = \vec{F} \cdot \vec{d}\). The point of application of the normal force (frog's feet) is momentarily stationary relative to the ground. Thus, the work done by normal reaction on the frog is zero. (A) is strictly false.

Reason (R): The normal reaction force is an external force. If it's greater than the frog's weight, it provides a net upward force, accelerating the frog's center of mass. So, (R) is true.

Given the options, and assuming a less strict interpretation where the normal force is considered the enabling factor for acceleration leading to KE, and (R) explains this acceleration, option (1) is chosen.

Since no external horizontal force acts on the system (boat + persons), the position of the centre of mass of the system remains unchanged. Thus, the shift in the centre of mass is zero. Option (c) is correct.

In gravity-free space, no external force acts, so the centre of mass remains at its initial height of $10\text{ m}$. Using COM conservation: $M_m h_m + M_s h_s = (M_m + M_s) Y_{cm} \implies 50(h) + 0.5(0) = (50 + 0.5)(10) \implies h = 10.1\text{ m}$. Option (b) is correct.

Concept: Since external force on the system is zero, the acceleration of the center of mass is zero. Formula: $v_{cm} = \frac{\sum m_i v_i}{\sum m_i}$. Solution: Since the system starts from rest and only internal forces act, velocity of center of mass remains zero.

Concept: Shift in center of mass must be zero. Formula: $m_1 \Delta x_1 = m_2 \Delta x_2$. Solution: Substituting $\Delta x_1 = d$ gives $\Delta x_2 = \frac{m_1}{m_2}d$.