It is easy to find $F_1 = (-1, 0), F_2 = (1, 0)$.
Denote $P(x_0, y_0)$, $Q_1(x_1, y_1)$, $Q_2(x_2, y_2)$. By the given condition, it follows that
$$x_0, y_0 > 0, \quad y_1 < 0, \quad y_2 < 0.$$
By the definition of ellipse we get
$$|PF_1| + |PF_2| = |Q_1F_1| + |Q_1F_2| = |Q_2F_1| + |Q_2F_2| = 2\sqrt{2}.$$
Hence, the perimeters of $\triangle PF_1Q_2$ and $\triangle PF_2Q_1$ are both $l = 4\sqrt{2}$.
And since $|F_1F_2| = 2$,
$$r_1 = \frac{2S_{\triangle PF_1Q_2}}{l} = \frac{(y_0 - y_2) \cdot |F_1F_2|}{l} = \frac{y_0 - y_2}{2\sqrt{2}}.$$
Similarly, we can get $r_2 = \frac{y_0 - y_1}{2\sqrt{2}}$, so $r_1 - r_2 = \frac{y_1 - y_2}{2\sqrt{2}}$.
In the following, we will first find $y_1 - y_2$.

Fig. 11.1

The equation of line $PF_1$ is $x = \frac{(x_0 + 1)y}{y_0} - 1$. Substituting it into $\frac{x^2}{2} + y^2 = 1$ and rearranging it gives
$$\left( \frac{(x_0 + 1)^2}{2y_0^2} + 1 \right) y^2 - \frac{x_0 + 1}{y_0} y - \frac{1}{2} = 0.$$
Multiplying both sides by $2y_0^2$ and noticing that $x_0^2 + 2y_0^2 = 2$, we find that
$$(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.$$
The two roots of this equation are $y_0$ and $y_1$. By Vieta's formulas, we get $y_0y_1 = -\frac{y_0^2}{3+2x_0}$. Thus,
$$y_1 = -\frac{y_0}{3 + 2x_0}.$$
Similarly, we can get $y_2 = -\frac{y_0}{3 - 2x_0}$. Therefore,
$$y_1 - y_2 = \frac{y_0}{3 - 2x_0} - \frac{y_0}{3 + 2x_0} = \frac{4x_0y_0}{9 - 4x_0^2}.$$
Since $9 - 4x_0^2 = \frac{1}{2}x_0^2 + 9y_0^2 \ge 2\sqrt{\frac{1}{2}x_0^2 \cdot 9y_0^2} = 3\sqrt{2}x_0y_0$, there is
$$r_1 - r_2 = \frac{y_1 - y_2}{2\sqrt{2}} = \frac{\sqrt{2}x_0y_0}{9 - 4x_0^2} \le \frac{\sqrt{2}x_0y_0}{3\sqrt{2}x_0y_0} = \frac{1}{3},$$
where the equality sign holds when $\frac{1}{2}x_0^2 = 9y_0^2$ is required. Accordingly, $x_0 = \frac{3\sqrt{5}}{5}$, $y_0 = \frac{\sqrt{10}}{10}$.
Therefore, the maximum of $r_1 - r_2$ is $y_0 = \frac{1}{3}$. $\square$