This problem involves triangle ABC with the following details:

• $AB = 8$,
• $AC = 5$,
• $\angle BAC = 60^\circ$, and we are also concerned with the circumcircle (外接円) of this triangle, denoted as $K$.

Let's solve it step by step:

(1) Finding $BC$

Using the Law of Cosines for $\triangle ABC$: $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$ Substitute the given values: $BC^2 = 8^2 + 5^2 - 2 \cdot 8 \cdot 5 \cdot \cos(60^\circ)$ Since $\cos(60^\circ) = \frac{1}{2}$, this simplifies to: $BC^2 = 64 + 25 - 80 \cdot \frac{1}{2} = 64 + 25 - 40 = 49$ $BC = \sqrt{49} = 7$ Thus, $BC = 7$.

(2) Finding the radius $r$ of the circumcircle $K$

The formula for the radius $r$ of the circumcircle is: $r = \frac{ABC}{s}$ where $ABC$ is the area of the triangle and $s$ is the semi-perimeter.

First, calculate the semi-perimeter $s$: $s = \frac{AB + AC + BC}{2} = \frac{8 + 5 + 7}{2} = 10$

Now, calculate the area of the triangle $ABC$. Using the formula for the area based on two sides and the included angle: $\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle BAC)$ Substitute the values: $\text{Area} = \frac{1}{2} \cdot 8 \cdot 5 \cdot \sin(60^\circ) = \frac{1}{2} \cdot 8 \cdot 5 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}$

Finally, the radius $r$ is: $r = \frac{AB \cdot AC \cdot BC}{4 \cdot \text{Area}} = \frac{8 \cdot 5 \cdot 7}{4 \cdot 10\sqrt{3}} = \frac{280}{40\sqrt{3}} = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}$ Thus, $K$'s radius is $\frac{7\sqrt{3}}{3}$.

(3) Determining the length of segment $CP$

The problem involves placing point $P$ on $BC$ such that $BP = 3$. We are asked to find the length of segment $CP$.

Since $BC = 7$ and $BP = 3$, we have: $CP = BC - BP = 7 - 3 = 4$

Final Answers:

1. $BC = 7$
2. $K$'s radius is $\frac{7\sqrt{3}}{3}$
3. $CP = 4$

Would you like further details or explanations on any part?

Here are 5 related questions to explore further:

1. What is the general form of the Law of Cosines and how can it be applied in other problems?
2. How is the circumradius related to the area of a triangle?
3. Can we derive the circumradius for a right triangle without using trigonometry?
4. What is the significance of the semi-perimeter in triangle geometry?
5. How can we apply Heron’s formula to find the area of $\triangle ABC$ instead?

Tip: Whenever using trigonometric formulas, be mindful of angle units (degrees vs. radians) to avoid errors in calculations.