3. Let the triangle: the sides opposite to the interior angles be $a$, $b$, $c$. If $a^{2}=b(b+c)$, $b^{2}=c(c+a)$, then, which of the following is true?
(A) $\angle A=2 \angle B=4 \angle C$
(B) $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$
(C) $\cos C-\cos B-\cos A=\frac{1}{2}$
(D) $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=2$
3. (D).

There are multiple methods to prove: $a^{2}=b(b+c) \Rightarrow \angle A=2 \angle B$.
Below is a plane geometry proof.
As shown in the figure, extend $C A$ to $D$,

such that $A D=A B$, and connect $B D$. Then
$C D=b+c$. Also, from $a^{2}=b(b+c)$ we get
$$
\frac{B C}{A C}=\frac{D C}{B C}.
$$

Therefore, $\triangle A B C \sim \triangle B C D$, which implies $\angle D=\angle A B C$.
$$
\therefore \angle B A C=\angle A B D+\angle A D B=2 \angle A B C.
$$

Similarly, from $b^{2}=c(c+a)$ we can get $\angle B=2 \angle C$. Thus,
$$
\angle A=2 \angle B=4 \angle C \text{. ((A) holds) }
$$

From $a^{2}=b(b+c)=b c+b^{2}$
$$
\begin{array}{l}
=b c+c(c+a)=a c+c(b+c) \\
=a c+c \cdot \frac{a^{2}}{b},
\end{array}
$$

Dividing both sides by $a^{2} c$, we get
$$
\frac{1}{c}=\frac{1}{a}+\frac{1}{b} \text{. ((B) holds) }
$$

From (1), we can solve for
$$
\begin{array}{l}
\angle C=\frac{\pi}{7}, \angle B=\frac{2 \pi}{7}, \angle A=\frac{4 \pi}{7} \text{. } \\
\therefore \cos C-\cos B-\cos A \\
=\frac{2 \sin \frac{3 \pi}{7}}{2 \sin \frac{3 \pi}{7}}\left(\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}\right) \\
=\frac{1}{2 \sin \frac{3 \pi}{7}}\left[\left(\sin \frac{4 \pi}{7}+\sin \frac{2 \pi}{7}\right)\right. \\
\left.-\left(\sin \frac{5 \pi}{7}+\sin \frac{\pi}{7}\right)+\sin \frac{6 \pi}{7}\right] \\
=\frac{\sin \frac{4 \pi}{7}}{2 \sin \frac{3 \pi}{7}}=\frac{1}{2} \text{. ((C) holds) } \\
\end{array}
$$

From (1), (2), and (3), we know that (D) should be selected.
In fact, from $\angle A=\frac{4 \pi}{7}>\frac{\pi}{2}$, we know that $\triangle A B C$ is an obtuse triangle, $\therefore \cos A<0$. Therefore,
$$
\begin{array}{l}
\sin ^{2} A+\sin ^{2} B+\sin ^{2} C \\
=\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}+\frac{1-\cos 2 C}{2} \\
=\frac{3}{2}-\frac{1}{2}(\cos 2 A+\cos 2 B+\cos 2 C) \\
=\frac{3}{2}+\frac{1}{2}\left(1+4 \cos A \cos B \cos \left(C^{\circ}\right)\right. \\
=2+4 \cos A \cos B \cos C<2.
\end{array}
$$