Problem 215

Given the ellipse and hyperbola equations, we can find the focal points $F_1$ and $F_2$. For the ellipse given by $\frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$, the standard form is $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$ where $a^2 = 6$ and $b^2 = 2$. The focal distance $c$ is found using $c^2 = a^2 - b^2 = 6 - 2 = 4$, so $c = 2$. Because the ellipse is centered at the origin and its major axis is along the x-axis, the foci are at $F_1(-2,0)$ and $F_2(2,0)$.

To find the point of intersection $P(x, y)$, we need to solve the system of equations:
$$\begin{cases} \frac{x^{2}}{6} + \frac{y^{2}}{2} = 1 \\ \frac{x^{2}}{3} - y^{2} = 1 \end{cases}$$

Multiplying the second equation by 2, we get:
$$\begin{cases} x^{2} + 3y^{2} = 6 \\ 2x^{2} - 6y^{2} = 6 \end{cases}$$

Adding them up yields:
$$3x^{2} - 3y^{2} + x^{2} + 3y^{2} = 6 + 6 \Rightarrow 4x^{2} = 12 \Rightarrow x^{2} = 3$$

Thus, $x = \pm\sqrt{3}$. Back substituting this into one of the original equations, we find that $y^{2} = \frac{1}{2}$, and thus $y = \pm\frac{\sqrt{2}}{2}$.

Out of the four possible points given by the pair of coordinates ($\pm\sqrt{3}$, $\pm\frac{\sqrt{2}}{2}$), we can select $P$ as $\left(\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ without any loss of generality.

Now, we find the vectors $\overrightarrow{PF_1}$ and $\overrightarrow{PF_2}$ as follows:
$$\overrightarrow{PF_1} = \left(-2 - \frac{3\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right), \quad \overrightarrow{PF_2} = \left(2 - \frac{3\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$$

The cosine of the angle between the vectors is given by:
$$\cos\angle F_1PF_2 = \frac{\overrightarrow{PF_1} \cdot \overrightarrow{PF_2}}{\left\|\overrightarrow{PF_1}\right\| \cdot \left\|\overrightarrow{PF_2}\right\|} = \frac{\left(-2 - \frac{3\sqrt{2}}{2}\right)\left(2 - \frac{3\sqrt{2}}{2}\right) + \left(-\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{2}}{2}\right)}{\sqrt{\left(-2 - \frac{3\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} \cdot \sqrt{\left(2 - \frac{3\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2}}$$

Simplifying the above expression, we get:
$$\cos\angle F_1PF_2 = \frac{1}{3}$$

Therefore, the solution is $\boxed{\frac{1}{3}}$.