Answer
$\dfrac{m}{n^{3/8}}$
Work Step by Step
Using laws of exponents, then,
\begin{array}{l}
\dfrac{(m^2n)^{1/4}}{m^{-1/2}n^{5/8}}
\\\\=
\dfrac{m^{2\cdot\frac{1}{4}}n^{\frac{1}{4}}}{m^{\frac{-1}{2}}n^{\frac{5}{8}}}
\\\\=
\dfrac{m^{\frac{2}{4}}n^{\frac{1}{4}}}{m^{\frac{-1}{2}}n^{\frac{5}{8}}}
\\\\=
m^{\frac{2}{4}-\frac{-1}{2}}n^{\frac{1}{4}-\frac{5}{8}}
\\\\=
m^{\frac{2}{4}+\frac{2}{4}}n^{\frac{2}{8}-\frac{5}{8}}
\\\\=
m^{\frac{4}{4}}n^{-\frac{3}{8}}
\\\\=
\dfrac{m}{n^{3/8}}
.\end{array}
