And $c$ is a constant), the difference in eccentricity between $C_{1}$ and $C_{2}$ does not exceed 1, and $C_{2}$ has an asymptote with a slope no less than $\frac{4}{3}$,
$C_{1}$ and $C_{2}$ intersect the positive $x$-axis at points $A$ and $B$, respectively, and the intersection point of the two curves in the first quadrant is $D$. Try to find the maximum area of $\triangle A B D$ and the equations of $C_{1}$ and $C_{2}$ at this time.
According to the problem, we can set the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, with eccentricity $e_{1}$, and the hyperbola $C_{2}$: $\frac{x^{2}}{m^{2}}-\frac{y^{2}}{n^{2}}=1(m>0, n>0)$, with eccentricity $e_{2}$. Given that $\frac{n}{m} \geqslant \frac{4}{3}$, we have $e_{2}=\sqrt{\frac{m^{2}+n^{2}}{m^{2}}} \geqslant \sqrt{1+\left(\frac{4}{3}\right)^{2}}=\frac{5}{3}$. Rewriting the equations of $C_{1}$ and $C_{2}$, we get:
$$
\left(1-e_{1}^{2}\right) x^{2}+y^{2}=\frac{1-e_{1}^{2}}{e_{1}^{2}} c^{2},\left(1-e_{2}^{2}\right) x^{2}+y^{2}=\frac{1-e_{2}^{2}}{e_{2}^{2}} c^{2},
$$

By solving the system of equations and eliminating $x$, we obtain $y=y_{D}=\frac{c}{e_{1} e_{2}} \sqrt{\left(1-e_{1}^{2}\right)\left(e_{2}^{2}-1\right)}$, and thus $S_{\triangle A B D}=$ $\frac{1}{2}\left(\frac{c}{e_{1}}-\frac{c}{e_{2}}\right) y_{D}=\frac{1}{2} c^{2} \cdot \frac{e_{2}-e_{1}}{e_{1}^{2} e_{2}^{2}} \sqrt{\left(1-e_{1}^{2}\right)\left(e_{2}^{2}-1\right)}$. Given that $e_{2}-e_{1} \leqslant 1$, we have $e_{1} \in \left[e_{2}-1,1\right), e_{2} \in \left[\frac{5}{3}, 2\right)$, so
$$
\begin{aligned}
S_{\triangle A B D} & \leqslant \frac{1}{2} c^{2} \cdot \frac{1}{\left(e_{2}-1\right)^{2} e_{2}^{2}} \sqrt{\left[1-\left(e_{2}-1\right)^{2}\right]\left(e_{2}^{2}-1\right)} \\
& =\frac{1}{2} c^{2} \sqrt{\frac{\left(2-e_{2}\right)\left(e_{2}+1\right)}{e_{2}^{3} \cdot\left(e_{2}-1\right)^{3}}},
\end{aligned}
$$

Let $f\left(e_{2}\right)=\frac{\left(2-e_{2}\right)\left(e_{2}+1\right)}{e_{2}^{3}\left(e_{2}-1\right)^{3}}$, then $f\left(e_{2}\right)=\frac{-e_{2}^{2}+e_{2}+2}{\left(e_{2}^{2}-e_{2}\right)^{3}}=\frac{-\left(e_{2}-\frac{1}{2}\right)^{2}+\frac{9}{4}}{\left[\left(e_{2}-\frac{1}{2}\right)^{2}-\frac{1}{4}\right]^{3}}$ is a decreasing function on $e_{2} \in \left[\frac{5}{3}, 2\right)$. Therefore,
$$
S_{\triangle A B D} \leqslant \frac{1}{2} c^{2} \sqrt{f\left(\frac{5}{3}\right)}=\frac{1}{2} c^{2} \cdot \frac{9 \sqrt{5}}{25}=\frac{9 \sqrt{5}}{50} c^{2},
$$

Thus, the maximum value of $S_{\triangle A B D}$ is $\frac{9 \sqrt{5}}{50} c^{2}$. At this point, $e_{2}=\frac{5}{3}, e_{1}=\frac{2}{3}$, the ellipse $C_{1}: \frac{x^{2}}{\frac{9}{4} c^{2}}+\frac{y^{2}}{\frac{5}{4} c^{2}}=1$, and the hyperbola $C_{2}: \frac{x^{2}}{\frac{9}{25} c^{2}}-\frac{y^{2}}{\frac{16}{25} c^{2}}=1$.