Đáp án:

\(\begin{array}{l}
1)\left\{ \begin{array}{l}
x = 3\\
y = 5
\end{array} \right.\\
2)a)\left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
b)m = \dfrac{9}{4}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)DK:x \ge  - 1\\
\left\{ \begin{array}{l}
x + y = 8\\
x + y + \sqrt {x + 1}  = 10
\end{array} \right.\\
 \to \left\{ {x + y + \sqrt {x + 1}  - x - y = 10 - 8} \right.\\
 \to \left\{ \begin{array}{l}
\sqrt {x + 1}  = 2\\
x + y = 8
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x + 1 = 4\\
x + y = 8
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 3\\
y = 5
\end{array} \right.\\
2)Thay:x = 2\\
Pt \to 4 - 5.2 + m - 2 = 0\\
 \to m = 8\\
Thay:m = 8\\
Pt \to {x^2} - 5x + 6 = 0\\
 \to \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
b)Xét:\Delta  > 0\\
 \to 25 - 4\left( {m - 2} \right) > 0\\
 \to \dfrac{{25}}{4} > m - 2\\
 \to \dfrac{{33}}{4} > m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 5\\
{x_1}{x_2} = m - 2
\end{array} \right.\\
Do:2\left( {\dfrac{1}{{\sqrt {{x_1}} }} + \dfrac{1}{{\sqrt {{x_2}} }}} \right) = 3\\
 \to \dfrac{1}{{\sqrt {{x_1}} }} + \dfrac{1}{{\sqrt {{x_2}} }} = \dfrac{3}{2}\\
 \to \dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \dfrac{2}{{\sqrt {{x_1}{x_2}} }} = \dfrac{9}{4}\\
 \to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} + \dfrac{2}{{\sqrt {{x_1}{x_2}} }} = \dfrac{9}{4}\\
 \to \dfrac{5}{{m - 2}} + \dfrac{2}{{\sqrt {m - 2} }} - \dfrac{9}{4} = 0\left( 2 \right)\left( {m \ne 2} \right)\\
Đặt:\sqrt {m - 2}  = t\left( {t \ge 0} \right)\\
\left( 2 \right) \to 5{t^2} + 2t - \dfrac{9}{4} = 0\\
 \to \left[ \begin{array}{l}
t = \dfrac{1}{2}\\
t =  - \dfrac{9}{{10}}\left( l \right)
\end{array} \right.\\
 \to \sqrt {m - 2}  = \dfrac{1}{2}\\
 \to m - 2 = \dfrac{1}{4}\\
 \to m = \dfrac{9}{4}
\end{array}\)