Explanation

(a)
We have, ${(1.002)}^{12}$ or it can be rewritten as ${(1 + 0.002)}^{12}$
$\Longrightarrow \ {(1.002)}^{12}$=$1 + \ ^{12}C_{1}(0.002) + \ ^{12}C_{2}{(0.002)}^{2} + \ ^{12}C_{3}{(0.002)}^{3} + ...$
We want the answer upto 4 decimal places and as such, we have left further expansion.
$\therefore\ (1.002)^{12} = 1 + 12(0.002) + \frac{12 \bullet 11}{1 \bullet 2}(0.002)^{2}$ $+ \frac{12 \bullet 11 \bullet 10}{1 \bullet 2 \bullet 3}{(0.002)}^{3} + ...$
$= 1 + 0.024 + 2.64 \times 10^{- 4} + 1.76 \times 10^{- 6} + ...$
$= 1.0242$