1. Given points $A(3,1), B\left(\frac{5}{3}, 2\right)$, and the four vertices of $\square A B C D$ are all on the graph of the function $f(x)=\log _{2} \frac{a x+b}{x-1}$. Then the area of $\square A B C D$ is $\qquad$
$-1 . \frac{26}{3}$.
Given $f(3)=1, f\left(\frac{5}{3}\right)=2$
$\Rightarrow a=1, b=1$.
Thus, $f(x)=\log _{2} \frac{x+1}{x-1}$.
Clearly, $f(x)$ is an odd function, and the graph of $f(x)$ is symmetric about the origin $O$.

From the given, $l_{A B}: 3 x+4 y-13=0$, the distance from point $O$ to $l_{A B}$ is $d=\frac{|-13|}{\sqrt{3^{2}+4^{2}}}=\frac{13}{5}$.
$$
\begin{array}{l}
\text { Also }|A B|=\frac{5}{3} \Rightarrow S_{\triangle A O B}=\frac{1}{2}|A B| d=\frac{13}{6} \\
\Rightarrow S_{\triangle A B C D}=\frac{13}{6} \times 4=\frac{26}{3} .
\end{array}
$$