Sure, here is the translated text:

```
II. Fill-in-the-Blanks (9 points each, total 54 points)
1. Let the three interior angles of $\triangle A B C$ be $\angle A, \angle B, \angle C$ with the corresponding side lengths $a, b, c$. If $a < b < c$, and
$$
\left\{\begin{array}{l}
\frac{b}{a}=\frac{\left|b^{2}+c^{2}-a^{2}\right|}{b c}, \\
\frac{c}{b}=\frac{\left|c^{2}+a^{2}-b^{2}\right|}{c a}, \\
\frac{a}{c}=\frac{\left|a^{2}+b^{2}-c^{2}\right|}{a b},
\end{array}\right.
$$

then the radian measures of $\angle A, \angle B, \angle C$ are in the ratio $\qquad$ .
```
II. 1.1:2:4.
From equation (1) and the Law of Sines and Cosines, we get
$$
\frac{\sin B}{\sin A}=2|\cos A| \Rightarrow \sin B=|\sin 2 A| \text {. }
$$

Since \(0<\angle A<\angle B<\frac{\pi}{2}\), we have \(\angle B=2 \angle A\).
Similarly, from equation (2), we get \(\angle C=2 \angle B\).
Combining \(\angle A+\angle B+\angle C=\pi\), we get \(\angle A=\frac{\pi}{7}, \angle B=\frac{2 \pi}{7}, \angle C=\frac{4 \pi}{7}\).
From equation (3), we get \(\sin A=|\sin 2 C|\).
Clearly, \(\angle A=\frac{\pi}{7}\) and \(\angle C=\frac{4 \pi}{7}\) satisfy the above equation.
Therefore, \(\angle A: \angle B: \angle C=1: 2: 4\).