Answer
$a)$ $\dfrac{x^{4}}{y}$
$b)$ $\dfrac{8y^{8}}{x^{2}}$
Work Step by Step
$a)$ $\Big(\dfrac{x^{3/2}}{y^{-1/2}}\Big)^{4}\Big(\dfrac{x^{-2}}{y^{3}}\Big)$
First, evaluate the power:
$\Big(\dfrac{x^{3/2}}{y^{-1/2}}\Big)^{4}\Big(\dfrac{x^{-2}}{y^{3}}\Big)=\Big(\dfrac{x^{6}}{y^{-2}}\Big)\Big(\dfrac{x^{-2}}{y^{3}}\Big)=...$
Evaluate the product of fractions and simplify if possible:
$...=\dfrac{x^{6+(-2)}}{y^{-2+3}}=\dfrac{x^{4}}{y}$
$b)$ $\Big(\dfrac{4y^{3}z^{2/3}}{x^{1/2}}\Big)^{2}\Big(\dfrac{x^{-3}y^{6}}{8z^{4}}\Big)^{1/3}$
Evaluate both powers:
$\Big(\dfrac{4y^{3}z^{2/3}}{x^{1/2}}\Big)^{2}\Big(\dfrac{x^{-3}y^{6}}{8z^{4}}\Big)^{1/3}=\Big(\dfrac{16y^{6}z^{4/3}}{x}\Big)\Big(\dfrac{x^{-1}y^{2}}{2z^{4/3}}\Big)=...$
Evaluate the product:
$...=\dfrac{16x^{-1}y^{6+2}z^{4/3}}{2xz^{4/3}}=\dfrac{16x^{-1}y^{8}z^{4/3}}{2xz^{4/3}}=...$
Evaluate the division and simplify if possible:
$...=8x^{-1-1}y^{8}z^{4/3-4/3}=8x^{-2}y^{8}=\dfrac{8y^{8}}{x^{2}}$
