\({x^2} + ax + b + 2 = 0\).

Ta có \(\Delta = {a^2} - 4\left( {b + 2} \right) = {a^2} - 4b - 8\).

Để phương trình có 2 nghiệm phân biệt thì \(\Delta > 0 \Leftrightarrow {a^2} - 4b - 8 > 0\) (*).

Khi đó, áp dụng định lí Vi-ét ta có: \(\left\{ \begin{array}{l}{x_1} + {x_2} = - a\\{x_1}{x_2} = b + 2\end{array} \right.\).

Theo bài ra ta có:

\(\begin{array}{l}\left\{ \begin{array}{l}{x_1} - {x_2} = 4\\x_1^3 - x_2^3 = 28\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1} - {x_2} = 4\\{\left( {{x_1} - {x_2}} \right)^3} + 3{x_1}{x_2}\left( {{x_1} - {x_2}} \right) = 28\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x_1} - {x_2} = 4\\{4^3} + 12{x_1}{x_2} = 28\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1} - {x_2} = 4\\{x_1}{x_2} = - 3\end{array} \right.\end{array}\)

Mà \({x_1}{x_2} = b + 2 \Rightarrow b + 2 = - 3 \Leftrightarrow b = - 3 - 2 = - 5\).

Ta có : \(\left\{ \begin{array}{l}{x_1} + {x_2} = - a\\{x_1} - {x_2} = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{x_1} = 4 - a\\2{x_2} = - a - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1} = \frac{{4 - a}}{2}\\{x_2} = \frac{{ - a - 4}}{2}\end{array} \right.\)

\(\begin{array}{l} \Rightarrow {x_1}{x_2} = - 3 \Leftrightarrow \frac{{4 - a}}{2}.\left( {\frac{{ - a - 4}}{2}} \right) = - 3\\ \Leftrightarrow \left( {4 - a} \right)\left( {a + 4} \right) = 12\\ \Leftrightarrow 16 - {a^2} = 12\\ \Leftrightarrow {a^2} = 4 \Leftrightarrow \left[ \begin{array}{l}a = 2\\a = - 2\end{array} \right..\end{array}\)

Với \({a^2} = 4,\,\,b = - 5 \Rightarrow {a^2} - 4b - 8 = 4 - 4.\left( { - 5} \right) - 8 = 16 > 0 \Rightarrow \) thỏa mãn điều kiện \(\left( * \right)\).

Vậy có 2 cặp số \(\left( {a;b} \right)\) thỏa mãn yêu cầu bài toán là \(\left( {a;b} \right) = \left( {2; - 5} \right)\) hoặc \(\left( {a;b} \right) = \left( { - 2; - 5} \right)\).

\( \cdot x_1^3-x_2^3=28\\↔(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=28\\↔4(x_1^2+x_1x_2+x_2^2)=28\\↔x_1^2+x_1x_2+x_2^2=7\\\\\cdot x_1-x_2=4\\↔(x_1-x_2)^2=16\\↔x_1^2-2x_1x_2+x_2^2=16\\↔7-3x_1x_2=16\\↔3x_1x_2=-9\\↔x_1x_2=-3\\\\\cdot x_1^2+x_1x_2+x_2^2=7\\↔(x_1+x_2)^2-x_1x_2=7\\↔(x_1+x_2)^2+3=7\\↔(x_1+x_2)^2=4\\↔\left[\begin{array}{1}x_1+x_2=2\\x_1+x_2=-2\end{array}\right.\\Suy\,\,ra:\left[\begin{array}{1}\begin{cases}x_1+x_2=-2\\x_1x_2=-3\end{cases}\\\begin{cases}x_1+x_2=2\\x_1x_2=-3\end{cases}\end{array}\right.\\Theo\,\,Vi-et\,\,đảo:\\Pt→\left[\begin{array}{1}x^2+2x-3=0\\x^2-2x-3=0\end{array}\right.\\→\left[\begin{array}{1}\begin{cases}a=2\\b+2=-3\end{cases}\\\begin{cases}a=-2\\b+2=-3\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}a=2\\b=-5\end{cases}\\\begin{cases}a=-2\\b=-5\end{cases}\end{array}\right.\)

Vậy \( (a;b)=\{ (2;-5);(-2;-5)\}\)