8. In Rt $\triangle A C B$, $\angle C=90^{\circ}, D, E$ are on $B C, C A$ respectively, and $\frac{B D}{A C}=\frac{A E}{C D}=K$, let $B E \cap A D=$ $O$, if $\angle B O D=75^{\circ}$, then $K=$ $\qquad$ .
11. It is easy to observe that: $x=1$ is a natural number root of the original cubic equation. By synthetic division, the original cubic equation can be reduced to a quadratic equation $5 x^{2}-5 p x+66 p-1=0$.

The three roots of the original cubic equation being natural numbers is equivalent to the two roots of the quadratic equation (1) being natural numbers.

Let $u, v(u\frac{66}{5}$, i.e., $u \geqslant 14$.

Since 2 does not divide $u$ and 3 does not divide $u$, we have $u \geqslant 17$. By $v \geqslant u$ and (3), we get $\frac{66 u-1}{5 u-66} \geqslant u$,

i.e., $5 u^{2}-132 u+1 \leqslant 0$. Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<$ $\frac{132}{5}$,

so $17 \leqslant u \leqslant 26$. Furthermore, since 2 does not divide $u$ and 3 does not divide $u$, $u$ can only be $17,19,23,25$.
When $u=17$, by (3) we get $v=\frac{1121}{19}=59$.
When $u=19$, by (3) we get $v=\frac{1253}{29}$, which is not a natural number, so it should be discarded.
When $u=23$, by (3) we get $v=\frac{1517}{49}$, which is not a natural number, so it should be discarded.
When $u=25$, by (3) we get $v=\frac{1649}{59}$, which is not a natural number, so it should be discarded.
Therefore, only when $p=u+v=76$, the two roots of equation (1) are natural numbers, and the three roots of the original equation are natural numbers.
8. Solution: Construct rectangle $A C D P$, and connect $P A, P E, P D, P B$.
$$
\begin{array}{l}
\because \frac{B D}{A C}=\frac{A E}{C D}=k, \therefore \frac{B D}{D P}=\frac{A E}{A P}=k, \\
\because \mathrm{Rt} \triangle B D P \backsim \mathrm{Rt} \triangle E A P, \\
\therefore \frac{A P}{P E}=\frac{D P}{P B}, \angle A P D=\angle B P E, \\
\therefore \triangle A P D \sim \triangle E P B, \\
\therefore \angle B O D=\angle D P B \text{ (the rotation angle equals the angle between corresponding sides) } \\
\because k=\tan \angle D P B=\tan \angle B O D=2+\sqrt{3} .
\end{array}
$$