All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let $A_{50}$ and $B_{50}$ denote the total time A & B take respectively, in seconds.

$A_{50} = \frac{50}{2}[2T+49(2)] = 50T + 2450$

$1.5H = 5400s$

$1.75H = 6300s$

$5400 \le 50T +2450 \le 6300$

$59 \le T \le 77$

$\therefore, \{ T\in \mathbb{R} | 59 \le T \le 77\}$

(ii)
$B_{50} = \frac{t(1.02^{50}-1)}{1.02-1} = 50t(1.02^{50}-1)$

$5400 \le 50t(1.02^{50}-1) \le 6300$

$63.84 \le t \le 74.483$

$63.9 \le t \le 74.4$ (3 SF)

$\therefore, \{ t\in \mathbb{R} | 63.9 \le t \le 74.4\}$

(iii)
For A: Time taken $= 59 + 49(2) = 157$
For B: Time taken $= 63.84533(1.02)^{49} = 168.475$

Difference $= 168.475 - 157 \approx 11 \mathrm{seconds}$ (nearest seconds)