8. If the equation $\frac{|x|}{x+4}=k x^{2}$ has four distinct real solutions for $x$, then the range of values for $k$ is
8. $k \in\left(\frac{1}{4},+\infty\right)$.

Obviously, $x=0$ is a solution to the given equation.
When $x \neq 0$, the given equation becomes
$$
\frac{1}{x+4}=k|x| \text {. }
$$

We need to find that $\frac{1}{x+4}=k|x|$ has three distinct non-zero real solutions, which means the graphs of $y=\frac{1}{x+4}$ and $y=k|x|$ intersect at three points.
(1) $k>0$.
When $x>0$, $y=k|x|$ and $y=\frac{1}{x+4}$ intersect at one point;

When $x<0$, for $k>\frac{1}{4}$, the graphs of $y=k|x|$ and $y=\frac{1}{x+4}$ intersect at two points, satisfying the condition.
(2) $k=0$.
The equation $\frac{|x|}{x+4}=k x^{2}$ has only one real solution, which does not meet the requirement.
(3) $k<0$.
The graphs of $y=k|x|$ and $y=\frac{1}{x+4}$ intersect at only one point, which does not meet the requirement.
In summary, $k \in\left(\frac{1}{4},+\infty\right)$.