Dựa vào đồ thị hàm số \(y={f}'\left( 1+x \right)\) ta có \({f}'\left( 1+x \right)=0\).

\(\Leftrightarrow \left[ \begin{align} & x=0 \\ & x=1 \\ & x=2 \\ \end{align} \right.\)

Đặt \(t=1+x\Rightarrow {f}'\left( t \right)=0.\)

\(\Leftrightarrow \left[ \begin{align} & t=1 \\ & t=2 \\ & t=3 \\ \end{align} \right.\)

Vậy

\({f}'\left( t \right)\ge 0\Leftrightarrow \left[ \begin{align} & t\le 1 \\ & 2\le t\le 3 \\ \end{align} \right.\)\(\,,\,{f}'\left( t \right)\le 0\) \(\Leftrightarrow \left[ \begin{align} & 1\le t\le 2 \\ & t\ge 3 \\ \end{align} \right.\).

\(g\left( x \right)=f\left( -{{x}^{2}}+2x-2022+m \right)\)\( \Rightarrow {g}'\left( x \right)=\left( 2-2x \right){f}'\left( -{{x}^{2}}+2x-2022+m \right)\).

Hàm số \(g\left( x \right)=f\left( -{{x}^{2}}+2x-2022+m \right)\) đồng biến trên \(\left( 0\,;\,1 \right)\)\( \Leftrightarrow \) \(\left( 2-2x \right){f}'\left( -{{x}^{2}}+2x-2022+m \right)\ge 0\,,\,\forall x\in \left( 0;1 \right)\)\( \Leftrightarrow {f}'\left( -{{x}^{2}}+2x-2022+m \right)\ge 0\,,\,\forall x\in \left( 0;1 \right)\).

\( \Leftrightarrow \left[ \begin{array}{l} - {x^2} + 2x - 2022 + m \le 1\\ \left\{ \begin{array}{l} - {x^2} + 2x - 2022 + m \ge 2\\ - {x^2} + 2x - 2022 + m \le 3 \end{array} \right. \end{array} \right.{\kern 1pt} {\kern 1pt} \forall x \in \left( {0;1} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l} m \le {x^2} - 2x + 2023\\ \left\{ \begin{array}{l} m \ge {x^2} - 2x + 2024\\ m \le {x^2} - 2x + 2025 \end{array} \right. \end{array} \right.{\kern 1pt} {\kern 1pt} \forall x \in \left( {0;1} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l} m \le 2022\\ 2024 \le m \le 2024 \end{array} \right.\)

Vậy có \(2023\) số.