10. (This sub-question is worth 20 points) As shown in the figure, rectangle $A B C D$ is folded upwards along a line segment $E F$ parallel to $A D$ (point $E$ moves on line segment $A B$, and point $F$ moves on line segment $C D$), forming a triangular prism $A B E-C D F$. It is known that $A B=5, A C=\sqrt{34}$.
(1) If $\triangle A B G$ is a right triangle, where $G$ is a point on line segment $E F$, find the range of the length $x$ of line segment $E G$;
(2) If the length of $E G$ in (1) is the largest integer within its range, and the length of line segment $A B$ is minimized, find the value of the dihedral angle $C-E F-D$;
(3) Under the conditions of (1) and (2), find the volume of the tetrahedron $A-B F G$.
(1) From the given conditions, we know that $\triangle A E G$ and $\triangle B E G$ are both right triangles, so $A G^{2}=A E^{2}+x^{2}, B G^{2}=B E^{2}+x^{2}$. By the cosine rule: $A B^{2}=A E^{2}+B E^{2}-2 A E \cdot B E \cos \angle A E B$.
5 points
Thus,
$$
\begin{array}{l}
2 x^{2}+A E^{2}+B E^{2}=A B^{2}=A E^{2}+B E^{2}-2 A E \cdot B E \cos \angle A E B \\
x^{2}=-A E \cdot B E \cos \angle A E B < A E \cdot B E = t(5-t) = -t^{2} + 5 t \leq 2.5^{2}
\end{array}
$$

Therefore, $x \in [0, 2.5)$, and for
$$
\forall k \in [0, 2.5), \quad A E = E B = 2.5, \quad \angle A E B = \pi - \arccos \frac{k^{2}}{2.5^{2}}
$$

Then $x = \sqrt{-A E \cdot B E \cos \angle A E B} = k$, so the range of $x$ is $[0, 2.5)$.
10 points
(2) Since $A E \perp E F, B E \perp E F$, $\angle A E B$ is the plane angle of the dihedral angle $C-E F-D$. From (1), the length of $E G$ is the largest integer in $[0, 2.5)$, so $x = 2$. Thus:
$$
A B^{2} = t^{2} + (5-t)^{2} + 4 = 2 t^{2} - 10 t + 29, \quad t \in (0, 5)
$$

Therefore, when $t = 2.5$, the length of segment $A B$ is minimized, which gives: $2 = -\frac{25}{4} \cos \angle A E B, \angle A E B = \pi - \arccos \frac{8}{25}$. 15 points
(3) From (1) and (2), we know
$$
\angle A E B = \pi - \arccos \frac{8}{25}, A E = E B = \frac{5}{2}, A G = B G = \frac{\sqrt{41}}{2}, E G = 2
$$

And $E F = \sqrt{A C^{2} - A B^{2}} = \sqrt{34 - 25} = 3$.
Since $A E \perp E F, B E \perp E F, A E \cap B E = E$, $E F \perp$ plane $E A B$, so
$$
\begin{aligned}
V_{A-B F G} & = V_{A-B E F} - V_{A-B E G} = \frac{1}{3} \left( S_{\triangle A E B} E F - S_{\triangle A G B} E G \right) \\
& = \frac{1}{3} \left( \left( \frac{1}{2} A E^{2} \sin \angle A E B \right) E F - \frac{1}{2} B G^{2} E G \right) = \frac{1}{6} \left( \frac{25}{4} \sqrt{1 - \frac{64}{625}} \cdot 3 - \frac{41}{4} \cdot 2 \right) \\
& = \frac{3 \sqrt{561} - 82}{24}
\end{aligned}
$$
20 points