Ta có: \(2 x^{2}+3 x+2\gt 0, \forall x \in \mathbb{R}\) vì \(\left\{\begin{array}{l}\Delta=-7\lt 0 \\ a=2>0\end{array}\right.\).

Khi đó bất phương trình trở thành: \(-\left(2 x^{2}+3 x+2\right) \leq x^{2}-5 x+m\lt 7\left(2 x^{2}+3 x+2\right)\)

\(\Leftrightarrow\left\{\begin{array} { l } { - ( 2 x ^ { 2 } + 3 x + 2 ) \leq x ^ { 2 } - 5 x + m } \\{ x ^ { 2 } - 5 x + m \lt 7 ( 2 x ^ { 2 } + 3 x + 2 ) }\end{array} \Leftrightarrow \left\{\begin{array}{l}3 x^{2}-2 x+m+2 \geq 0 \\13 x^{2}+26 x-m+14>0\end{array}\right.\right.\)

Xét \((1):\) \(3 x^{2}-2 x+m+2 \geq 0, \forall x \in \mathbb{R} \Leftrightarrow\left\{\begin{array}{l}a=3>0\\ \Delta^{\prime}=1-3(m+2) \leq 0\end{array} \Leftrightarrow m \geq-\frac{5}{3}\right.\).

Xét \((2) :\) \(13 x^{2}+26 x-m+14>0, \forall x \in \mathbb{R} \Leftrightarrow\left\{\begin{array}{l}a=13>0 \\ \Delta^{\prime}=13^{2}-13(14-m)\lt 0\end{array} \Leftrightarrow m\lt 1\right.\).

Vậy \(m \in\left[-\frac{5}{3} ; 1\right) \xrightarrow{m \in \mathcal{Z}} m \in\{-1 ; 0\}\) nên có hai giá trị thỏa mãn.