In the ellipse $x^{2}+4 y^{2}=8$, $AB$ is a moving chord of length $\frac{5}{2}$, and $O$ is the origin. Find the range of the area of $\triangle AOB$.

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Given the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$, we can set $A(2 \sqrt{2} \cos \alpha, \sqrt{2} \sin \alpha), B(2 \sqrt{2} \cos \beta, \sqrt{2} \sin \beta), \alpha, \beta \in[0,2 \pi)$. Then,
$$
\begin{aligned}
|A B|^{2} & =\frac{25}{4}=8(\cos \alpha-\cos \beta)^{2}+2(\sin \alpha-\sin \beta)^{2} \\
& =6\left(\cos ^{2} \alpha+\cos ^{2} \beta\right)-16 \cos \alpha \cos \beta-4 \sin \alpha \sin \beta+4,
\end{aligned}
$$

Thus, $3(\cos 2 \alpha+\cos 2 \beta)-16 \cos \alpha \cos \beta-4 \sin \alpha \sin \beta+\frac{15}{4}=0$, so $6 \cos (\alpha+\beta) \cos (\alpha-\beta)-8(\cos (\alpha+\beta)+\cos (\alpha-\beta))+2(\cos (\alpha-\beta)-\cos (\alpha-\beta))+\frac{15}{4}=0$, which leads to $\cos (\alpha-\beta)=\frac{6 \cos (\alpha+\beta)-\frac{15}{4}}{6 \cos (\alpha+\beta)-10}$. Given $\cos (\alpha+\beta) \in[-1,1]$, we have $\cos (\alpha-\beta) \in\left[-\frac{9}{16}, \frac{39}{64}\right]$, and thus $|\sin (\alpha-\beta)| \in\left[\frac{5 \sqrt{103}}{64}, 1\right]$.

The equation of the line $A B$ is $(\sin \alpha-\sin \beta) x-2(\cos \alpha-\cos \beta) y+2 \sqrt{2} \sin (\beta-\alpha)=0$. Therefore, the distance from point $O$ to $l_{A B}$ is
$$
d=\frac{2 \sqrt{2}|\sin (\alpha-\beta)|}{(\sin \alpha-\sin \beta)^{2}+1(\cos \alpha-\cos \beta)^{2}}=\frac{2 \sqrt{2}|\sin (\alpha-\beta)|}{\sqrt{\frac{25}{8}}}=\frac{8}{5}|\sin (\alpha-\beta)|,
$$

Thus,
$$
\begin{aligned}
S_{\triangle A O B} & =\frac{1}{2} d|A B|=\frac{1}{2} \cdot \frac{8}{5}|\sin (\alpha-\beta)| \cdot \frac{5}{2} \\
& =2|\sin (\alpha-\beta)| \in\left[\frac{5 \sqrt{103}}{32}, 2\right] .
\end{aligned}
$$

When the slope $k$ of the line $A B$ exists, let $l_{A B}: y=k x+m$. Solving the system of equations $\left\{\begin{array}{l}y=k x+m, \\ x^{2}+4 y^{2}=8,\end{array}\right.$, we get: $\left(4 k^{2}+1\right) x^{2}+8 k m x+4\left(m^{2}-2\right)=0$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}\right)$,
$$
\begin{aligned}
\frac{25}{4} & =|A B|^{2}=\left(1+k^{2}\right)\left(x_{1}-x_{2}\right)^{2} \\
& =\left(1+k^{2}\right) \frac{64 k^{2} m^{2}-4 \cdot\left(4 k^{2}+1\right) \cdot 4\left(m^{2}-2\right)}{\left(4 k^{2}+1\right)^{2}} \\
& -\frac{\left(1+k^{2}\right)\left[32\left(4 k^{2}+1\right)-16 m^{2}\right]}{\left(4 k^{2}+1\right)^{2}}
\end{aligned}
$$

Simplifying, we get $m^{2}=2\left(4 k^{2}+1\right)-\frac{25}{64} \frac{\left(4 k^{2}+1\right)^{2}}{k^{2}+1}$. The distance from the origin $O$ to $l_{A B}$ is $d=\frac{|m|}{\sqrt{1+k^{2}}}$, so
$$
S_{\triangle A O B}^{2}=\frac{1}{4} d^{2} \cdot|A B|^{2}=\frac{25}{16} \frac{m^{2}}{1+k^{2}}=\frac{25}{16}\left[2 \frac{4 k^{2}+1}{1+k^{2}}-\frac{25}{64} \cdot\left(\frac{4 k^{2}+1}{1+k^{2}}\right)^{2}\right]
$$

Let $t=\frac{4 k^{2}+1}{1+k^{2}}$; then $t=4-\frac{3}{1+k^{2}} \in[1,4)$, thus
$$
S_{\triangle A O B}^{2}=\frac{25}{16}\left(2 t-\frac{25}{64} t^{2}\right)=-\frac{625}{1024} t^{2}+\frac{25}{8} t=-\frac{625}{1024}\left(t-\frac{64}{25}\right)^{2}+1,
$$

Hence, $S_{\triangle A O B}^{2} \in\left[\frac{2575}{1024}, 4\right]$, which means $S_{\triangle A O B} \in\left[\frac{5 \sqrt{103}}{32}, 2\right]$.
When the slope $k$ of the line $A B$ does not exist, $S_{\triangle A O B}=\frac{1}{2} \cdot \frac{\sqrt{7}}{2} \cdot \frac{5}{2}=\frac{5 \sqrt{7}}{8} \in\left[\frac{5 \sqrt{103}}{32}, 2\right]$. In summary, $S_{\triangle A O B} \in\left[\frac{5 \sqrt{103}}{32}, 2\right]$.