[ Height of a pyramid (tetrahedron).]

The base of the pyramid is a right triangle with a hypotenuse of 6 and an acute angle of \(15^{\circ}\). All lateral edges are inclined to the plane of the base at an angle of \(45^{\circ}\). Find the volume of the pyramid.
Let the angle at vertex $C$ of the right triangle $A B C$ be $90^{\circ}$, and the angle at vertex $A$ be $15^{\circ}$. Since the lateral edges $D A, D B$, and $D C$ of the pyramid $A B C D$ form equal angles with the plane of the base, the height $D O$ of the pyramid passes through the center of the circle circumscribed around the right triangle $A B C$, i.e., through the midpoint $O$ of the hypotenuse $A B$. Then $D A O$ is the angle of the lateral edge with the plane of the base of the pyramid. According to the problem, $\angle D A O=45^{\circ}$. From the right triangle $A O D$, we find that $D O=O A=\frac{1}{2} A B=\frac{1}{2} 6=3$.

Let $C K$ be the height of the right triangle $A B C$. The angle $K O C$ is the external angle of the isosceles triangle $A O C$, so

$$
\angle K O C=2 \cdot \angle O A C=2 \cdot 15^{\circ}=30^{\circ}
$$

From the right triangle $K O C$, we find that

$$
C K=\frac{1}{2} O C=\frac{1}{2} \frac{1}{2} A B=\frac{1}{2} \frac{1}{2} 6=3 / 2
$$

Therefore,

$$
V(A B C D)=\frac{1}{3} S(A B C) \cdot D O=\frac{1}{3} \frac{1}{2} A B \cdot C K \cdot D O=\frac{1}{3} \frac{1}{2} 6 \cdot(3 / 2) \cdot 3=9 / 2
$$

Answer

4.5.