8. Given non-negative real numbers $u, v, w$ satisfying $u+v+w=2$. Then the range of $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$ is $\qquad$ .
8. $[0,1]$.

Obviously, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2} \geqslant 0$, and when $u=0, v=0$, $w=0$, the equality holds.
Therefore, its minimum value is 0.
Assume without loss of generality that $u=\max \{u, v, w\}>0$.
Then $v w \leqslant u^{2}<2 u^{2} \Rightarrow v w-2 u^{2}<0$
$\Rightarrow v w\left(v w-2 u^{2}\right) \leqslant 0$.
Thus, $u^{2} v^{2}+v^{2} w^{2}+w^{2} u^{2}$
$=u^{2}(v+w)^{2}+v w\left(v w-2 u^{2}\right)$
$\leqslant[u(v+w)]^{2} \leqslant\left[\frac{(u+v+w)^{2}}{4}\right]^{2}=1$.
When $u=v=1, w=0$, the equality holds in the above expression.
Therefore, the maximum value is 1.
Hence, the required range of values is $[0,1]$.