8. Given a tetrahedron $ABCD$ where $AB=CD=2a$, $AC=BD=BC=AD=\sqrt{10}$, then the range of values for $a$ is $\qquad$.
8. Take the midpoint $O$ of $CD$, then $|AO|=|BO|=\sqrt{|AC|^{2}-|CO|^{2}}=\sqrt{10-a^{2}}$. In $\triangle AOB$, by the cosine rule, $\cos \angle AOB=\frac{|AO|^{2}+|BO|^{2}-|AB|^{2}}{2 \cdot|AO| \cdot|BO|}=\frac{20-6a^{2}}{2(10-a^{2})}=\frac{10-3a^{2}}{10-a^{2}}$. Since, $0<\cos \angle AOB<1$, we have $0<\frac{10-3a^{2}}{10-a^{2}}<1$, which implies $0<a<\sqrt{5}$. Therefore, the answer is $0<a<\sqrt{5}$.