II. (16 points) Let the side lengths opposite to two interior angles of $\triangle A B C$ be $a, b, c$ respectively, and $a+b+c=16$. Find
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$

the value.
$$
\begin{array}{l} 
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 4 R^{2}\left(\sin ^{2} B \cdot \cos ^{2} \frac{C}{2}+\sin ^{2} C \cdot \cos ^{2} \frac{B}{2}+\right. \\
\left.2 \sin B \cdot \sin C \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\right. \\
\left.2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\
= 16 R^{2} \cos \frac{B}{2} \cdot \cos \frac{C}{2}\left(\frac{1-\cos B}{2}+\right. \\
\left.\frac{1-\cos C}{2}+\frac{\cos A+\cos B+\cos C-1}{2}\right) \\
= 16 R^{2}\left(\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)^{2} \\
= R^{2}(\sin A+\sin B+\sin C)^{2} \\
=\left(\frac{a+b+c}{2}\right)^{2}=64 .
\end{array}
$$

Solution 2: As shown in Figure 2, extend line segment $BC$ to points $E$ and $F$ such that $CE = AC = b$ and $BF = AB = c$. Connect $AE$ and $AF$. Let the midpoints of $AE$ and $AF$ be $M$ and $N$, respectively. Connect $BN$, $CM$, and $NM$.
$$
\begin{array}{l}
\text { Then } AM = b \cos \frac{C}{2}, AN = c \cos \frac{B}{2}, \\
\angle MAN = \angle A + \frac{1}{2}(\angle B + \angle C) = \frac{\pi}{2} + \frac{\angle A}{2}. \\
\text { Therefore, } b^{2} \cos ^{2} \frac{C}{2} + c^{2} \cos ^{2} \frac{B}{2} + 2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= AM^{2} + AN^{2} - 2 AM \cdot AN \cos \angle MAN \\
= MN^{2} = \left(\frac{a+b+c}{2}\right)^{2} = 64 .
\end{array}
$$