5. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, points $X$ and $Y$ are the centers of the squares $A A_{1} B_{1} B$ and $B B_{1} C_{1} C$, respectively, and point $Z$ is on the diagonal $B D$ such that $D Z=3 Z B$. Then the area of the section cut by the plane $X Y Z$ from the circumscribed sphere of the cube is $\qquad$.
5. $\frac{5 \pi}{8}$.

From the problem, we know that the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is cut by a plane $X Y Z$ parallel to $B B_{1}$, and the cross-section is a rectangle. Let $O$ be the center of the cube (i.e., the center of the circumscribed sphere).
Then the distance from point $O$ to the plane $X Y Z$ = the distance from point $O$ to the line $X Y$
$$
=B Z=\frac{1}{4} B D=\frac{\sqrt{2}}{4} \text {. }
$$

The radius of the circumscribed sphere is $O B=\frac{\sqrt{3}}{2}$, thus, the radius $r$ of the circle formed by the intersection of the plane $X Y Z$ with the circumscribed sphere satisfies
$$
r^{2}=\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{2}}{4}\right)^{2}=\frac{5}{8} \text {. }
$$

Therefore, the required area is $\frac{5 \pi}{8}$.