Đáp án:

$\begin{array}{l}
1)\\
{x^2} - 2\left( {m + 1} \right)x - 3 = 0\\
 \Rightarrow \Delta ' > 0\\
 \Rightarrow {\left( {m + 1} \right)^2} + 3 > 0\left( {luon\,dung} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} =  - 3
\end{array} \right.\\
\left| {{x_1}} \right| - \left| {{x_2}} \right| = 5\\
 \Rightarrow x_1^2 - 2\left| {{x_1}} \right|.\left| {{x_2}} \right| + x_2^2 = 25\\
 \Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} - 2.\left| { - 3} \right| = 25\\
 \Rightarrow 4{\left( {m + 1} \right)^2} - 2.\left( { - 3} \right) - 2.3 = 25\\
 \Rightarrow {\left( {m + 1} \right)^2} = \dfrac{{25}}{4}\\
 \Rightarrow \left[ \begin{array}{l}
m =  - 1 + \dfrac{5}{2} = \dfrac{3}{2}\\
m =  - 1 - \dfrac{5}{2} = \dfrac{{ - 7}}{2}
\end{array} \right.\\
2)\\
{x^2} - 6x + m = 0\\
 \Rightarrow \Delta ' > 0\\
 \Rightarrow 9 - m > 0\\
 \Rightarrow m < 9\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1}{x_2} = m
\end{array} \right.\\
x_1^2 - x_2^2 = 12\\
 \Rightarrow \left( {{x_1} + {x_2}} \right)\left( {{x_1} - {x_2}} \right) = 12\\
 \Rightarrow {x_1} - {x_2} = 2\\
 \Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 4\\
 \Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 4\\
 \Rightarrow 36 - 4.m = 4\\
 \Rightarrow m = 8\left( {tmdk} \right)\\
3)\\
\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 4} \right)\left( {x + 5} \right) = 10\\
 \Rightarrow \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right)\left( {x + 4} \right) = 10\\
 \Rightarrow \left( {{x^2} + 6x + 5} \right)\left( {{x^2} + 6x + 8} \right) = 10\\
Dat:{x^2} + 6x + 5 = a\\
 \Rightarrow a.\left( {a + 3} \right) = 10\\
 \Rightarrow {a^2} + 3a - 10 = 0\\
 \Rightarrow \left( {a - 2} \right)\left( {a + 5} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
a - 2 = 0\\
a + 5 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} + 6x + 3 = 0\\
{x^2} + 6x + 10 = 0\left( {vô\,nghiệm} \right)
\end{array} \right.\\
 \Rightarrow {x^2} + 6x + 9 - 6 = 0\\
 \Rightarrow {\left( {x + 3} \right)^2} = 6\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 3 + \sqrt 6 \\
x =  - 3 - \sqrt 6 
\end{array} \right.
\end{array}$