Đáp án:

$\begin{array}{l}
a){x^2} + mx - 16 = 0\\
 \Rightarrow \Delta  > 0\\
 \Rightarrow {\left( { - m} \right)^2} - 4.\left( { - 16} \right) > 0\\
 \Rightarrow {m^2} + 64 > 0\left( {\text{luôn}\,\text{đúng}} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} =  - \dfrac{b}{a} =  - m\\
{x_1}{x_2} = \dfrac{c}{a} =  - 16
\end{array} \right.\\
Khi:{x_1} =  - 4{x_2}\\
 \Rightarrow  - 4{x_2}.{x_2} =  - 16\\
 \Rightarrow x_2^2 = 4\\
 \Rightarrow \left[ \begin{array}{l}
{x_2} = 2 \Rightarrow {x_1} =  - 8\\
{x_2} =  - 2 \Rightarrow {x_1} = 8
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2 + \left( { - 8} \right) =  - m\\
\left( { - 2} \right) + 8 =  - m
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
m = 6\\
m =  - 6
\end{array} \right.\\
\text{Vậy}\,m =  \pm 6\\
b)m{x^2} + 2\left( {m - 4} \right)x + m + 7 = 0\\
 + Dk:m \ne 0\\
\Delta ' > 0\\
 \Rightarrow {\left( {m - 4} \right)^2} - m\left( {m + 7} \right) > 0\\
 \Rightarrow {m^2} - 8m + 16 - {m^2} - 7m > 0\\
 \Rightarrow  - 15m + 16 > 0\\
 \Rightarrow m < \dfrac{{16}}{{15}}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{ - 2m + 8}}{m}\\
{x_1}{x_2} = \dfrac{{m + 7}}{m}
\end{array} \right.\\
{x_1} - 2{x_2} = 0\\
 \Rightarrow {x_1} = 2{x_2}\\
 \Rightarrow 3{x_2} = \dfrac{{ - 2m + 8}}{m}\\
 \Rightarrow {x_2} = \dfrac{{ - 2m + 8}}{{3m}}\\
 \Rightarrow {x_1} = \dfrac{{ - 4m + 16}}{{3m}}\\
 \Rightarrow {x_1}.{x_2} = \dfrac{{ - 2m + 8}}{{3m}}.\dfrac{{ - 4m + 16}}{{3m}}\\
 \Rightarrow \dfrac{{m + 7}}{m} = \dfrac{{8\left( {m - 4} \right)\left( {m - 4} \right)}}{{9{m^2}}}\\
 \Rightarrow m + 7 = \dfrac{{8{{\left( {m - 4} \right)}^2}}}{{9m}}\\
 \Rightarrow 9{m^2} + 63m = 8\left( {{m^2} - 16m + 64} \right)\\
 \Rightarrow 9{m^2} + 63m = 8{m^2} - 128m + 512\\
 \Rightarrow {m^2} + 191m - 512 = 0\\
 \Rightarrow \left[ \begin{array}{l}
m =  - 193,64\\
m = 2,6\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,m =  - 193,64
\end{array}$