Đáp án:

$\begin{array}{l}
1){x^2} - mx - 2 = 0\\
 \Rightarrow \Delta  > 0\\
 \Rightarrow {m^2} - 4.\left( { - 2} \right) > 0\\
 \Rightarrow {m^2} + 8 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} =  - 2
\end{array} \right.\\
{x_1} - 2{x_2} = 5\\
 \Rightarrow {x_1} = 5 + 2{x_2}\\
 \Rightarrow 5 + 2{x_2} + {x_2} = m\\
 \Rightarrow {x_2} = \dfrac{{m - 5}}{3}\\
 \Rightarrow {x_1} = 5 + 2{x_2} = \dfrac{{2m + 5}}{3}\\
 \Rightarrow \dfrac{{m - 5}}{3}.\dfrac{{2m + 5}}{3} =  - 2\\
 \Rightarrow 2{m^2} - 5m - 25 =  - 18\\
 \Rightarrow 2{m^2} - 5m - 7 = 0\\
 \Rightarrow \left( {2m - 7} \right)\left( {m + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
m = \dfrac{7}{2}\\
m =  - 1
\end{array} \right.\\
Vậy\,m =  - 1;m = \dfrac{7}{2}\\
B2)\\
\Delta ' > 0\\
 \Rightarrow {\left( {m + 1} \right)^2} - {m^2} - 2 > 0\\
 \Rightarrow {m^2} + 2m + 1 - {m^2} - 2 > 0\\
 \Rightarrow 2m - 1 > 0\\
 \Rightarrow m > \dfrac{1}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = {m^2} + 2
\end{array} \right.\\
x_1^2 + x_2^2 = 10\\
 \Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 10\\
 \Rightarrow 4{\left( {m + 1} \right)^2} - 2.\left( {{m^2} + 2} \right) = 10\\
 \Rightarrow 4{m^2} + 8m + 4 - 2{m^2} - 4 = 10\\
 \Rightarrow 2{m^2} + 8m - 10 = 0\\
 \Rightarrow {m^2} + 4m - 5 = 0\\
 \Rightarrow \left( {m - 1} \right)\left( {m + 5} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
m = 1\\
m =  - 5\left( {ktm} \right)
\end{array} \right.\\
Vậy\,m = 1
\end{array}$