Explanation

(a)
Let the term containing $x^{7}$ in the expansion of
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( {ax}^{2} + \frac{1}{bx} \right)^{8}is\ T_{r + 1}.$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{r + 1} = \ ^{8}C_{r}\left( ax^{2} \right)^{8 - r}\left( \frac{1}{bx} \right)^{r}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ ^{8}C_{r}\frac{a^{8 - r}}{b^{r}}x^{16 - 3r}$
Since, this term contains $x^{7}$.
$\therefore\ \ \ \ \ \ \ \ \ \ \ 16 - 3r = 7$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r = 3$
$\therefore Coefficient\ of\ x^{7}\ in\ the\ expansion\ of\ \left( {ax}^{2} + \frac{1}{bx} \right)^{8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ ^{8}C_{3} \bullet \frac{a^{5}}{b^{3}}$
$Also,\ the\ term\ containing\ x^{- 7}\ in\ the\ expansion\ of\ \left( ax - \frac{1}{{bx}^{2}} \right)^{8}is\ T_{R + 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{R + 1} = \ ^{8}C_{R}(a\ x)^{8 - R}\left( - \frac{1}{{bx}^{2}} \right)^{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = {( - 1)}^{R}\ ^{8}C_{R}\frac{a^{8 - R}}{b^{R}}x^{8 - 3R}$
Since, this term contains $x^{- 7}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ 8 - 3R = - 7$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 5$
$\therefore Coefficient\ of\ x^{- 7}\ in\ the\ expansion\ of\ \left( {a\ x}^{\ } - \frac{1}{{bx}^{2}} \right)^{8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( - 1)^{5}\ ^{8}C_{5} \bullet \frac{a^{3}}{b^{5}}$
According to the given condition,
$\ \ \ \ \ \ \ \left| \ ^{8}C_{3} \bullet \frac{a^{5}}{b^{3}} \right| = \left| \ ^{8}C_{5} \bullet \frac{a^{3}}{b^{5}} \right|$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ a^{2}b^{2} = 1 \Longrightarrow ab = 1$