Hui Shi $\left(-\frac{1}{2 e},-\frac{2}{e^{4}}\right)$.
Analysis Let $f(x)=x e^{-2 x}+k$, and $f^{\prime}(x)=(1-2 x) e^{-2 x}=0$, we get $x=\frac{1}{2}$.
When $x<\frac{1}{2}, f^{\prime}(x)>0, f(x)$ is an increasing function on $\left(-\infty, \frac{1}{2}\right)$; when $x>\frac{1}{2}$, $f^{\prime}(x)<0, f(x)$ is a decreasing function on $\left(\frac{1}{2}, +\infty\right)$. Therefore, $f(x)$ reaches its maximum at $x=\frac{1}{2}$, and $f\left(\frac{1}{2}\right)=\frac{1}{2 e}+k$.
Since $f(0)=k>0, \quad f(-2)=-2 e^{4}+k<0, \quad f(2)=2 e^{-4}+k<0$,
$$
x e^{-2 x}+k=0
$$
has exactly two real roots in the interval $(-2,2)$, hence the range of $k$ is $\left(-\frac{1}{2 e},-\frac{2}{e^{4}}\right)$.
3. Given the equation $x e^{-2 x}+k=0$ has exactly two real roots in the interval $(-2,2)$, then the range of values for $k$ is