In $\triangle A B C, A D$ and $B E$ are medians of $\triangle A B C$
$\begin{aligned}
A D & =\frac{7}{2} \Rightarrow \angle D A B=\frac{\pi}{8} \\
\angle A B E & =\frac{\pi}{4} \Rightarrow A O=\frac{2}{3} A D=\frac{2}{3} \times \frac{7}{2}=\frac{7}{3}
\end{aligned}$


$\angle A O B=\pi-\left(\frac{\pi}{8}+\frac{\pi}{4}\right)=\left(\pi-\frac{3 \pi}{8}\right)$
By Sine rule, $\frac{A O}{\sin \frac{\pi}{4}}=\frac{B O}{\sin \frac{\pi}{8}}$ $B O=\frac{\frac{7}{3} \times \sin \frac{\pi}{8}}{\sin \frac{\pi}{4}}=\frac{7}{3} \times \frac{\sin \frac{\pi}{8}}{2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}}=\frac{7}{6 \cos \frac{\pi}{8}}$
$\begin{aligned}
& \text { Area of } \triangle A O B=\frac{1}{2} A O \times B O \times \sin \angle A O B \\
& =\frac{1}{2} \times \frac{7}{3} \times \frac{7}{6} \times \frac{\sin \left(\pi-\frac{3 \pi}{8}\right)}{\cos \frac{\pi}{8}} \\
& =\frac{7}{36} \times \frac{\sin 3 \frac{\pi}{8}}{\cos \frac{\pi}{8}}=\frac{7}{36} \quad\left[\because \sin \frac{3 \pi}{8}=\cos \frac{3 \pi}{8}\right]
\end{aligned}$
Area of $\triangle A B C=3 \times \frac{7}{36}=\frac{7}{12}$