13. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the distance between the upper vertex $A$ and the left vertex $B$ is $\sqrt{41}$, and the eccentricity is $\frac{3}{5}$. Point $P(t, 0)$ is on the $x$-axis where $(-4 \leqslant t \leqslant-1)$.
(1) Find the equation of the ellipse;
(2) Connect $A P$ and let it intersect the ellipse at point $C$. Draw a vertical line from $C$ to the $x$-axis, intersecting the ellipse at another point $D$. Find the range of $S_{\triangle A B D}$.
(1) According to the problem, $\left\{\begin{array}{l}a^{2}+b^{2}=41, \\ \frac{c}{a}=\frac{3}{5}\end{array} \Rightarrow\left\{\begin{array}{l}a=5, \\ b=4 .\end{array}\right.\right.$ Therefore, the equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$.
(2) As shown in the figure, let $A P: x=-\frac{t}{4} y+t$,

Solving $\left\{\begin{array}{l}4 x=-t(y-4) \\ 16 x^{2}+25 y^{2}=400\end{array}\right.$
$$
\Rightarrow\left(t^{2}+25\right) y^{2}-8 t^{2} y+16 t^{2}-400=0 \text {. }
$$

Then $y_{A}-y_{D}=y_{A}+y_{C}=\frac{8 t^{2}}{t^{2}+25}$,
Also, $D\left(x_{C},-y_{C}\right)$, thus $A D: y=-\frac{y_{C}+4}{x_{C}} x+4 \Rightarrow E\left(\frac{4 x_{C}}{y_{C}+4}, 0\right)$.
Therefore, $|E B|=\left|x_{E}\right|-5=\frac{-4 x_{C}}{y_{C}+4}-5=\frac{t\left(y_{C}-4\right)}{y_{C}+4}-5=-\frac{25}{t}-5=-\frac{5(t+5)}{t}$,
Hence, $S_{\triangle A B D}=\frac{1}{2}|E B| \cdot\left|y_{A}-y_{D}\right|=\left|\frac{4 t^{2}}{t^{2}+25} \cdot \frac{5(t+5)}{-t}\right|=\left|\frac{20\left(t^{2}+5 t\right)}{t^{2}+25}\right|$.
Considering $f(t)=\frac{t^{2}+5 t}{t^{2}+25}, f^{\prime}(t)=\frac{(2 t+5)\left(t^{2}+25\right)-2 t\left(t^{2}+5 t\right)}{\left(t^{2}+25\right)^{2}}=\frac{-5\left(t^{2}-10 t-25\right)}{\left(t^{2}+25\right)^{2}}$, then $f(x)$ is decreasing on $(-4,5-5 \sqrt{2})$ and increasing on $(5-5 \sqrt{2},-1)$, so $f(x)_{\max }=f(-4)=-\frac{4}{41}, f(x)_{\min }=f(5-5 \sqrt{2})=\frac{1-\sqrt{2}}{2}$.
In conclusion, the range of $S_{\triangle A B D}$ is $\left[\frac{80}{41}, 10(\sqrt{2}-1)\right]$.