3. (i) (Grade 11) In $\triangle A B C$, the necessary and sufficient condition for
$$
\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2}
$$

to hold is ( ).
(A) $a c=b^{2}$
(B) $a^{2}+c^{2}=b^{2}$
(C) $a+c=2 b$
(D) $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$
(ii) (Grade 12) In square $A B C D$ with side length $2$, $E$ is the midpoint of side $A B$. If $\triangle E A D$ and $\triangle E B C$ are folded along $E D$ and $E C$ respectively, such that $E A$ and $E B$ coincide, forming a tetrahedron, then the volume of this tetrahedron is ( ).
(A) $\frac{1}{\sqrt{3}}$
(B) $\frac{2}{\sqrt{3}}$
(C) $\sqrt{3}$
(D) $\frac{4}{\sqrt{3}}$
3. (i) C.
$$
\begin{array}{l}
\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2} \\
\Leftrightarrow \frac{1-\cos A}{2}+\frac{1-\cos C}{2}=\cos B \\
\Leftrightarrow 1-\cos B=\frac{1}{2}(\cos A+\cos C) \\
\Leftrightarrow 2 \sin ^{2} \frac{B}{2}=\cos \frac{A+C}{2} \cdot \cos \frac{A-C}{2} \\
\Leftrightarrow 2 \sin \frac{B}{2}=\cos \frac{A-C}{2} \\
\Leftrightarrow 2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}=\cos \frac{B}{2} \cdot \cos \frac{A-C}{2} \\
\Leftrightarrow \sin B=\frac{1}{2}(\sin A+\sin C) \Leftrightarrow 2 b=a+c .
\end{array}
$$
(ii)A.

As shown in Figure 2, let the overlapping point of $A$ and $B$ after folding be $P$, then $\triangle P C D$ is an equilateral triangle. Since $E P \perp P C, E P \perp P D$, we know that $E P \perp$ plane $P C D$. Therefore,
$$
V=\frac{1}{3} P E \cdot S_{\triangle P C D}=\frac{1}{3} \times 1 \times \frac{\sqrt{3}}{4} \times 2^{2}=\frac{1}{\sqrt{3}} .
$$