Answer
$\textbf{(a)}\hspace{0.7cm}\dfrac{a^2}{4b^3}$
$\textbf{(b)}\hspace{0.9cm}\dfrac{25x^{4}}{y^2}$
$\textbf{(c)}\hspace{0.9cm}\dfrac{y^3}{18z^3}$
Work Step by Step
Use that:
(1) $(ab)^m=a^mb^m $
(2) $(a^m)^n=a^{mn} $
(3) $a^{-m}=\dfrac{1}{a^m} $
(4) $\dfrac{a^m}{a^n}=a^{m-n}, \hspace{0.5cm} a\ne0 $
(5) $a^m.a^n=a^{m+n} $
$\textbf{(a)}\hspace{0.9cm}\dfrac{\frac{1}{2}a^{-3}b^{-4}}{2a^{-5}b^{-1}}=\dfrac{\frac{1}{2}}{2}.\dfrac{a^{-3}}{a^{-5}}.\dfrac{b^{-4}}{b^{-1}}=\dfrac{1}{4}.a^{-3-(-5)}.b^{-4-(-1)}=\dfrac{1}{4}a^2b^{-3}=\dfrac{a^2}{4b^3}$
$\textbf{(b)}\hspace{0.9cm}\bigg(\dfrac{x^2y}{5x^4}\bigg)^{-2}=\bigg(\dfrac{5x^4}{x^2y}\bigg)^2=\dfrac{(5x^4)^2}{(x^2y)^2}=\dfrac{25x^8}{x^4y^2}=\dfrac{25x^{8-4}}{y^2}=\dfrac{25x^{4}}{y^2}$
$\textbf{(c)}\hspace{0.9cm}\bigg(\dfrac{2y^{-1}z}{z^2}\bigg)^{-1}\bigg(\dfrac{y}{3z^2}\bigg)^2=\bigg(2y^{-1}z^{-1}\bigg)^{-1}\bigg(\dfrac{y}{3z^2}\bigg)^2=\frac{1}{2}yz\bigg(\dfrac{y^2}{3^2(z^2)^2}\bigg)=\frac{1}{2}\bigg(\dfrac{y^3}{9z^3}\bigg)=\dfrac{y^3}{18z^3}$
