5. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=$ $2, A A_{1}=A D=1$, points $E$, $F$, and $G$ are the midpoints of edges $A A_{1}$, $C_{1} D_{1}$, and $B C$ respectively. The volume of the tetrahedron $B_{1}-E F G$ is $\qquad$
5. $\frac{3}{8}$.

Solution: Take a point $H$ on the extension line of $D_{1} A_{1}$ such that $A_{1} H = \frac{1}{4}$. It is easy to prove that $H E \parallel B_{1} G$, 
$H E \parallel$ plane $B_{1} F G$. Therefore, $V_{B_{1}-E F G}=V_{H B_{1} F G}=$ $V_{H B_{1} F G}=V_{G B_{1} F H}$.

And $S_{\triangle B_{1} F H}=\frac{9}{8}$, the distance from $G$ to plane $B_{1} F H$ is 1. Hence, $V_{B_{1}-E F G}=\frac{3}{8}$.