10. Given the ellipse $C: \frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$ with an eccentricity of $\frac{1}{2}$, the upper and lower foci are $F_{1}, F_{2}$, and the right vertex is $D$. A line perpendicular to $D F_{2}$ through $F_{1}$ intersects the ellipse $C$ at points $A, B$, and $|B D|-\left|A F_{1}\right|=\frac{8 \sqrt{3}}{39}$.
(1) Find the value of $|A D|+|B D|$.
(2) Construct the tangents to the ellipse $C$ at points $A, B$ which intersect at point $E$. If $F_{1} E$ intersects the $x$-axis at $P$ and $F_{2} E$ intersects the $x$-axis at $Q$, find the value of $|P Q|$.
(1) $a=2 c, b=\sqrt{3} c$, connecting $F_{1} D$, then
$\triangle F_{1} F_{2} D$ is an equilateral triangle $\Rightarrow|O M|=\frac{\sqrt{3} c}{3}$
Thus, $A B: x=-\frac{\sqrt{3}}{3} y+\frac{\sqrt{3} c}{3}$,
Solving $\left\{\begin{array}{l}\sqrt{3} x=-y+c, \\ 3 y^{2}+4 x^{2}=12 c^{2}\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow 13 y^{2}-8 c y-32 c^{2}=0 \\
\Rightarrow|A B|=\sqrt{1+\frac{1}{3}} \cdot \frac{\sqrt{64 c^{2}+4 \cdot 13 \cdot 32 c^{2}}}{13} \\
=\frac{2 \sqrt{3}}{3} \cdot \frac{24 \sqrt{3} c}{13}=\frac{48 c}{13} .
\end{array}
$$

Then, $|B D|-\left|A F_{1}\right|=\left|B F_{2}\right|-\left|A F_{1}\right|=2 a-\left(\left|B F_{1}\right|+\left|A F_{1}\right|\right)=4 c-|A B|=\frac{4 c}{13}=\frac{8 \sqrt{3}}{39}$
$\Rightarrow c=\frac{2 \sqrt{3}}{3} \Rightarrow a=\frac{4 \sqrt{3}}{3}, b=2$. Therefore, $|A D|+|B D|=\left|A F_{2}\right|+\left|B F_{2}\right|=4 a-\left(\left|A F_{1}\right|+\left|B F_{1}\right|\right)$
$$
=4 a-|A B|=\frac{8 \sqrt{3}}{3}+\frac{8 \sqrt{3}}{39}=\frac{112 \sqrt{3}}{39} \text {. }
$$
(2) Let $E\left(x_{0}, y_{0}\right), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, from (1) we know the equation of the ellipse $C$ is $3 y^{2}+4 x^{2}=16$,

Then $A E: 3 y_{1} y+4 x_{1} x=16, B E: 3 y_{2} y+4 x_{2} x=16 \Rightarrow A B: 3 y_{0} y+4 x_{0} x=16$.
And $A B: x=-\frac{\sqrt{3}}{3} y+\frac{2}{3} \Rightarrow 3 x+\sqrt{3} y=2 \Rightarrow 8 \sqrt{3} y+24 x=16$,
Thus $y_{0}=\frac{8 \sqrt{3}}{3}, x_{0}=6 \Rightarrow E\left(6, \frac{8 \sqrt{3}}{3}\right)$. Also, $F_{1}\left(0, \frac{2 \sqrt{3}}{3}\right), F_{2}\left(0,-\frac{2 \sqrt{3}}{3}\right)$,
Therefore $k_{E F_{1}}=\frac{\frac{8 \sqrt{3}}{3}-\frac{2 \sqrt{3}}{3}}{6}=\frac{\sqrt{3}}{3} \Rightarrow E F_{1}: y=\frac{\sqrt{3}}{3} x+\frac{2 \sqrt{3}}{3} \Rightarrow P(-2,0)$,
$k_{E F_{2}}=\frac{\frac{8 \sqrt{3}}{3}+\frac{2 \sqrt{3}}{3}}{6}=\frac{5 \sqrt{3}}{9} \Rightarrow E F_{2}: y=\frac{5 \sqrt{3}}{9} x-\frac{2 \sqrt{3}}{3} \Rightarrow Q\left(\frac{6}{5}, 0\right)$, so $|P Q|=\frac{6}{5}+2=\frac{16}{5}$.