6. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=5, A A_{1}=4, A D=3$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}$, and $B C$, respectively. Then the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$
$\begin{array}{l}\text { 6. } \frac{45}{16} \text { Detailed Explanation: Extend } D_{1} A_{1} \text { to } H_{1} \text { such that } A_{1} H=\frac{3}{4} \therefore H E / / B_{1} G, H E / / \text { plane } B_{1} F G, \\ \therefore V_{B_{1}-E F G}=V_{E-B_{1} F G}=V_{H-B_{1} F G}=V_{G-B_{1} F H} \text {. Also, } \because V_{G-B_{1} F H}=\frac{1}{3} \times 1 \times S_{\triangle F H B_{1}}, \\ S_{\triangle F H B_{1}}=\left(3+\frac{3}{4}\right) \times 5-\frac{1}{2} \times \frac{5}{2} \times\left(3+\frac{3}{4}\right)-\frac{1}{2} \times \frac{5}{2} \times 3-\frac{1}{2} \times \frac{3}{4} \times 5=\frac{135}{16}, \therefore V_{G-B_{1} F H}=\frac{45}{16} .\end{array}$