11. (13 points) In a non-isosceles $\triangle ABC$, the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively, and they satisfy
$(2c - b) \cos C = (2b - c) \cos B$.
(1) Find the size of $\angle A$;
(2) If $a = 4$, find the range of the area of $\triangle ABC$.
(1) Notice,
$$
\begin{array}{l}
(2 c-b) \cos C=(2 b-c) \cos B \\
\Rightarrow(2 \sin C-\sin B) \cos C \\
\quad=(2 \sin B-\sin C) \cos B \\
\Rightarrow \sin 2 C-\sin 2 B=\sin (B-C) \\
\Rightarrow 2 \cos (C+B) \cdot \sin (C-B)=\sin (B-C) .
\end{array}
$$

Since $\triangle A B C$ is not an isosceles triangle, we have
$$
\sin (B-C) \neq 0 \text {. }
$$

Thus, $\cos (C+B)=-\frac{1}{2}$
$$
\begin{array}{l}
\Rightarrow \angle C+\angle B=120^{\circ} \\
\Rightarrow \angle A=60^{\circ} .
\end{array}
$$
(2) By the Law of Sines, we get
$$
\begin{array}{l} 
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{8}{\sqrt{3}} \\
\Rightarrow b=\frac{8}{\sqrt{3}} \sin B, c=\frac{8}{\sqrt{3}} \sin C \\
\Rightarrow S=\frac{1}{2} b c \sin A=\frac{16}{\sqrt{3}} \sin B \cdot \sin C \\
= \frac{16}{\sqrt{3}} \sin B \cdot \sin \left(120^{\circ}-B\right) \\
= \frac{8 \sqrt{3}}{3} \sin \left(2 B-30^{\circ}\right)+\frac{4 \sqrt{3}}{3} .
\end{array}
$$

Since $\angle B \in\left(0,120^{\circ}\right)$, we have $S \in(0,4 \sqrt{3})$.