(3) Let the two foci of the ellipse $\frac{x^{2}}{2}+y^{2}=1$ be $F_{1}$ and $F_{2}$. Draw a chord $AB$ through the right focus $F_{2}$ with an inclination angle of $\frac{\pi}{4}$. Then the area of $\triangle ABF_{1}$ is $\qquad$.
(3) $\frac{4}{3}$ Hint: Given $F_{1}(-1,0), F_{2}(1,0)$, and the equation of $AB$ is $x=y+1$,
substituting into the ellipse equation yields $\frac{(y+1)^{2}}{2}+y^{2}=1$, which simplifies to $3 y^{2}+2 y-1=0$. Let the roots of this equation be $y_{1}, y_{2}$, then
$$
\begin{aligned}
\left|y_{1}\right|+\left|y_{2}\right| & =\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}} \\
& =\sqrt{\left(-\frac{2}{3}\right)^{2}+\frac{4}{3}}=\frac{4}{3}
\end{aligned}
$$

Therefore,
$$
S_{A B F_{1}}=\frac{1}{2}\left|F_{1} F_{2}\right| \cdot\left(\left|y_{1}\right|+\left|y_{2}\right|\right)=\frac{4}{3} .
$$