Explanation

(a) On a planet, if a body dropped initial velocity $(u = 0)$ from a height $h$ and takes time $t$ to reach the ground then $h = \frac{1}{2}g_{P}t^{2} \Rightarrow g_{P} = \frac{2h}{t^{2}} = \frac{2 \times 8}{4} = 4\ m/s^{2}$
Using$\ T = 2\pi\sqrt{\frac{l}{g}} \Rightarrow T = 2\pi\sqrt{\frac{1}{4}} = \pi = 3.14\ sec$