Đáp án:

$\begin{array}{l}
Dkxd:1 \le x \le 3\\
Dat:\left\{ \begin{array}{l}
a = \sqrt {x - 1} \left( {a \ge 0} \right)\\
b = \sqrt {3 - x} \left( {b \ge 0} \right)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - 1 = {a^2}\\
3 - x = {b^2}
\end{array} \right. \Leftrightarrow \left\{ {{a^2} + {b^2} = 2} \right.\\
 \Leftrightarrow \left( {x - 1} \right).\left( {3 - x} \right) =  - {x^2} + 4x - 3\\
 =  - \left( {{x^2} - 4x + 6} \right) + 3\\
 \Leftrightarrow {x^2} - 4x + 6 = 3 - ab\\
Pt:\sqrt {x - 1}  + \sqrt {3 - x}  = {x^2} - 4x + 6\\
 \Leftrightarrow a + b = 3 - ab\\
 \Leftrightarrow \left\{ \begin{array}{l}
a + b = 3 - ab\\
{a^2} + {b^2} = 2
\end{array} \right.\\
 \Leftrightarrow {\left( {3 - ab} \right)^2} - 2ab = 2\\
 \Leftrightarrow {a^2}{b^2} - 6ab + 9 - 2ab = 2\\
 \Leftrightarrow {a^2}{b^2} - 8ab + 7 = 0\\
 \Leftrightarrow \left( {ab - 1} \right)\left( {ab - 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
ab = 1 \Leftrightarrow {x^2} - 4x + 6 = 3 - ab = 2\\
ab = 7 \Leftrightarrow {x^2} - 4x + 6 =  - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4x + 4 = 0 \Leftrightarrow x = 2\left( {tm} \right)\\
{x^2} - 4x + 10 = 0\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 2\\
b)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow 0 \le x \le 1\\
1 + \dfrac{2}{3}\sqrt {x - {x^2}}  = \sqrt x  + \sqrt {1 - x} \\
Dat:\sqrt x  + \sqrt {1 - x}  = a\\
 \Leftrightarrow x + 2\sqrt x .\sqrt {1 - x}  + 1 - x = {a^2}\\
 \Leftrightarrow 2\sqrt {x\left( {1 - x} \right)}  = {a^2} - 1\\
 \Leftrightarrow \dfrac{2}{3}\sqrt {x - {x^2}}  = \dfrac{{{a^2} - 1}}{3}\\
Pt \Leftrightarrow 1 + \dfrac{{{a^2} - 1}}{3} = a\\
 \Leftrightarrow 3 + {a^2} - 1 = 3a\\
 \Leftrightarrow {a^2} - 3a + 2 = 0\\
 \Leftrightarrow \left( {a - 1} \right)\left( {a - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 1 \Leftrightarrow \sqrt {x - {x^2}}  = \dfrac{{{a^2} - 1}}{2} = 0\\
a = 2 \Leftrightarrow \sqrt {x - {x^2}}  = \dfrac{{{a^2} - 1}}{2} = \dfrac{3}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 1\left( {tm} \right)\\
x - {x^2} = \dfrac{9}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1\\
{x^2} - x + \dfrac{9}{4} = 0\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 0;x = 1
\end{array}$