Answer
$\dfrac{a^{5/2}}{m^{23/12}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
\left( \dfrac{m^{-2/3}}{a^{-3/4}} \right)^4 \left( m^{-3/8}a^{1/4} \right)^{-2}
.$
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{m^{-\frac{2}{3}\cdot4}}{a^{-\frac{3}{4}\cdot4}} \right) m^{-\frac{3}{8}\cdot(-2)}a^{\frac{1}{4}\cdot(-2)}
\\\\=
\left( \dfrac{m^{-\frac{8}{3}}}{a^{-3}} \right) m^{\frac{3}{4}}a^{-\frac{1}{2}}
\\\\=
\dfrac{m^{-\frac{8}{3}}m^{\frac{3}{4}}a^{-\frac{1}{2}}}{a^{-3}}
.\end{array}
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{m^{-\frac{8}{3}+\frac{3}{4}}a^{-\frac{1}{2}}}{a^{-3}}
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
m^{-\frac{8}{3}+\frac{3}{4}}a^{-\frac{1}{2}-(-3)}
\\\\=
m^{-\frac{8}{3}+\frac{3}{4}}a^{-\frac{1}{2}+3}
.\end{array}
Changing the exponents to similar fractions, the expression above is equivalent to
\begin{array}{l}\require{cancel}
m^{-\frac{32}{12}+\frac{9}{12}}a^{-\frac{1}{2}+\frac{6}{2}}
\\\\=
m^{-\frac{23}{12}}a^{\frac{5}{2}}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{a^{\frac{5}{2}}}{m^{\frac{23}{12}}}
\\\\=
\dfrac{a^{5/2}}{m^{23/12}}
.\end{array}
