8. Given two triangles with side lengths of $3,4,5$, four triangles with side lengths of $4,5, \sqrt{41}$, and six triangles with side lengths of $\frac{5}{6} \sqrt{2}, 4,5$. Using these triangles as faces, we can form $\qquad$ tetrahedra.
When four identical triangles form a tetrahedron, it is called an isosceles tetrahedron. Since the isosceles tetrahedron is inscribed in a rectangular parallelepiped, its faces must be acute triangles. Clearly, the triangle with side lengths $4,5, \sqrt{41}$ is a right triangle, and the triangle with side lengths $\frac{5}{6} \sqrt{2}, 4,5$ is an obtuse triangle, so it is impossible to form a tetrahedron with four identical triangles given in the problem.
Since 4 and 5 are common sides of all triangles, they must be the lengths of two pairs of opposite edges of the tetrahedron, so the faces of the tetrahedron consist of two identical triangles each.

When two of the triangles have side lengths $4,5, \sqrt{41}$, the distance between the remaining two vertices is greater than $S C=D E$. Since $A E=\frac{16}{\sqrt{41}} \Rightarrow D E=$ $\frac{9}{\sqrt{41}}>\frac{5}{6} \sqrt{2}$, the side lengths of the other two triangles can only be $3,4,5$.
When two of the triangles have side lengths $3,4,5$, the distance between the remaining two vertices is greater than $3>\frac{5}{6} \sqrt{2}$, so the side lengths of the other two triangles can only be $4,5, \sqrt{41}$. Therefore, only one tetrahedron can be formed.