(N/A) Part $(i)$: Using Kepler's third law,$T^{2} = \frac{4 \pi^{2}}{G M_{m}} R^{3}$.
Rearranging for the mass of Mars $(M_{m})$: $M_{m} = \frac{4 \pi^{2} R^{3}}{G T^{2}}$.
Given $T = 7 \text{ hours } 39 \text{ minutes} = (7 \times 60 + 39) \times 60 \text{ s} = 27540 \text{ s}$ and $R = 9.4 \times 10^{6} \text{ m}$.
$M_{m} = \frac{4 \times (3.14)^{2} \times (9.4 \times 10^{6})^{3}}{6.67 \times 10^{-11} \times (27540)^{2}} \approx 6.48 \times 10^{23} \text{ kg}$.
Part $(ii)$: Using Kepler's third law for planetary orbits around the Sun: $\frac{T_{M}^{2}}{T_{E}^{2}} = \frac{R_{M}^{3}}{R_{E}^{3}}$.
Given $\frac{R_{M}}{R_{E}} = 1.52$ and $T_{E} = 365 \text{ days}$.
$T_{M} = T_{E} \times (1.52)^{3/2} = 365 \times 1.873 \approx 684 \text{ days}$.