12. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with an eccentricity of $\frac{1}{2}$, $A_{1}, A_{2}$ are the left and right vertices of the ellipse $C$, $F_{1}, F_{2}$ are the left and right foci of the ellipse $C$, $B$ is the upper vertex of the ellipse $C$, and the circumradius of $\triangle B A_{1} F_{1}$ is $\frac{2 \sqrt{21}}{3}$.
(1) Find the equation of the ellipse $C$;
(2) Suppose a line $l$ intersects the ellipse $C$ at points $P, Q$ (with $P, Q$ on opposite sides of the $x$-axis), and the slopes of the lines $A_{1} P, P A_{2}$, $A_{2} Q, Q A_{1}$ are denoted as $k_{1}, k_{2}, k_{3}, k_{4}$, respectively. Given that $k_{1}+k_{4}=\frac{5}{3}\left(k_{2}+k_{3}\right)$, find the range of the area of $\triangle F_{2} P Q$.
(1) $\frac{b}{a}=\sqrt{1-e^{2}}=\frac{\sqrt{3}}{2}$, let $\angle B A_{1} F_{1}=\theta$, then $2 R=\frac{a}{\sin \theta}=\frac{a \sqrt{a^{2}+b^{2}}}{b}$
$$
\Rightarrow \frac{4 \sqrt{21}}{3}=\frac{2 \sqrt{a^{2}+b^{2}}}{\sqrt{3}} \Rightarrow a^{2}+b^{2}=28 \Rightarrow a^{2}=16, b^{2}=12 \text {, }
$$

Thus, the equation of the ellipse $C$ is $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$.
(2) Let the line $P Q: y=k x+m, P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and solve the system of equations
$$
\left\{\begin{array}{l}
y=k x+m, \\
3 x^{2}+4 y^{2}=48
\end{array} \Rightarrow\left(4 k^{2}+3\right) x^{2}+8 k m x+4 m^{2}-48=0\right. \text {. }
$$

Then $k_{1}=\frac{y_{1}}{x_{1}+4}, k_{2}=\frac{y_{1}}{x_{1}-4}, k_{3}=\frac{y_{2}}{x_{2}-4}, k_{4}=\frac{y_{2}}{x_{2}+4}$, according to the problem, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}+4}+\frac{y_{2}}{x_{2}+4}=\frac{5}{3}\left(\frac{y_{1}}{x_{1}-4}+\frac{y_{2}}{x_{2}-4}\right) \\
\Rightarrow \frac{2 k x_{1} x_{2}+(m+4 k)\left(x_{1}+x_{2}\right)+8 m}{x_{1} x_{2}+4\left(x_{1}+x_{2}\right)+16}=\frac{5}{3} \cdot \frac{2 k x_{1} x_{2}+(m-4 k)\left(x_{1}+x_{2}\right)-8 m}{x_{1} x_{2}-4\left(x_{1}+x_{2}\right)+16} \\
\Rightarrow \frac{8 k\left(m^{2}-12\right)-8 k m(m+4 k)+8 m\left(4 k^{2}+3\right)}{4 m^{2}-48-32 k m+16\left(4 k^{2}+3\right)} \\
=\frac{5}{3} \cdot \frac{8 k\left(m^{2}-12\right)-8 k m(m-4 k)-8 m\left(4 k^{2}+3\right)}{4 m^{2}-48+32 k m+16\left(4 k^{2}+3\right)} \\
\Rightarrow \frac{-3(4 k-m)}{(4 k-m)^{2}}=\frac{-5(4 k+m)}{(4 k+m)^{2}} \Rightarrow \frac{3}{4 k-m}=\frac{5}{4 k+m} \Rightarrow k=m .
\end{array}
$$

Thus, the line $P Q$ passes through the fixed point $M(-1,0)$, and thus $S_{\triangle F_{2} P Q}=\frac{3}{2}\left|y_{1}-y_{2}\right|=\frac{3}{2}\left|k\left(x_{1}-x_{2}\right)\right|$
$$
=\frac{3}{2}\left|\frac{k \sqrt{64 k^{4}-4\left(4 k^{2}+3\right)\left(4 k^{2}-48\right)}}{4 k^{2}+3}\right|=\frac{18|k| \sqrt{5 k^{2}+4}}{4 k^{2}+3}
$$
$\xlongequal{k^{2}+\frac{3}{4}=t} \frac{9}{2} \sqrt{-\frac{3}{16} \cdot \frac{1}{t^{2}}-\frac{7}{2} \cdot \frac{1}{t}+5} \in\left(0, \frac{9 \sqrt{5}}{2}\right)$, when $\frac{1}{t}=\frac{4}{3}$, $S_{\triangle F_{2} P Q}=0$.
Therefore, the range of the area of $\triangle F_{2} P Q$ is $\left(0, \frac{9 \sqrt{5}}{2}\right)$.