Three. (Full marks 25 points) If the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}-x-2>0, \\
2 x^{2}+(5+2 k) x+5 k<0
\end{array}\right.
$$

has only the integer solution $x=-2$, find the range of the real number $k$.
Three, because $x=-2$ is a solution to the system of inequalities, substituting $x=-2$ into the second inequality yields
$$
(2 x+5)(x+k)=[2 \cdot(-2)+5](-2+k)2$. Therefore, $-k>-2>-\frac{5}{2}$,
which means the solution to the second inequality is $-\frac{5}{2}<x<-k$.
These two inequalities have only the integer solution $x=-2$, which should satisfy
(i) $\left\{\begin{array}{l}x2, \\ -\frac{5}{2}<x<-k, \\ x \text { is an integer. }\end{array}\right.$
For (i), since $k<2$, the only integer solution is $x=-2$. To meet the problem's requirements, (ii) should have no integer solutions, which means
$$
-2<-k \leqslant 3,-3 \leqslant k<2 \text {. }
$$

Combining (i) and (ii), we have $-3 \leqslant k<2$.
(Xiao Lian, compiled)