Given a positive integer $n(n \geqslant 5)$, let the rectangle $ABCD$ have side lengths $AB=2a, BC=2b(a>b)$. Points $P_{1}, P_{2}, \cdots, P_{n}=C, Q_{1}, Q_{2}, \cdots, Q_{n}=D$, and $R_{0}=D, R_{1}, R_{2}, \cdots, R_{n}=C$ are such that: $BP_{1}=P_{1}P_{2}=\cdots=P_{n-1}P_{n}=\frac{2b}{n}$, $AQ_{1}=Q_{1}Q_{2}=\cdots=Q_{n-1}Q_{n}=\frac{2b}{n}$, and $R_{0}R_{1}=R_{1}R_{2}=\cdots=R_{n-1}R_{n}=\frac{2a}{n}$.
(1) For $i=1,2, \cdots, n$, find the equation of the conic section $F_{i}$ on which the intersection points of $AP_{i}$ and $BR_{n-i}$, and $AR_{i}$ and $BQ_{i}$ lie;
(2) If points $S_{1}, S_{2}, \cdots, S_{n}$ on the extension of $DC$ satisfy $CS_{1}=S_{1}S_{2}=\cdots=S_{n-1}S_{n}=\frac{2a}{n}$, for $i=1,2, \cdots, n$, find the equation of the conic section $F_{2}$ on which the intersection points of $AP_{i}$ and $BS_{i}$ lie;
(3) Let $M$ be the point on the conic section $F_{1}$ that is farthest from $\Lambda B$. Find the minimum distance between $M$ and the points on the conic section $F_{2}$.
(1) Establish a Cartesian coordinate system with the base $AB$ as the $x$-axis and the perpendicular bisector of $AB$ as the $y$-axis. Let $A P_{i}$ and $B R_{n-i}$ intersect at point $E(x, y)$, and let the projections of $E, R_{n-i}$ on $AB$ be $F, G$, respectively. Then, $\triangle A E F \sim \triangle A P_{i} B, \triangle B E F \sim \triangle B R_{n} G$, so we have
$$
\frac{y}{a+x}=\frac{\frac{2 i b}{n}}{2 a}=\frac{i b}{n a}, \frac{y}{a-x}=\frac{2 b}{\frac{2 i a}{n}}=\frac{n b}{i a},
$$

Multiplying the two equations, we get the equation of the conic section $F_{1}$ that point $E$ satisfies: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Similarly, the intersection point of $A R_{i}$ and $B Q_{i}$ also lies on the conic section $F_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
(2) Let $A P_{i}$ and $B S_{i}$ intersect at point $H(x, y)$, and let the projections of $H, S_{i}$ on the line $AB$ be $K, L$, respectively. Then, $\triangle A B P_{i} \sim \triangle A K H, \triangle B H K \sim \triangle B S_{i} L$, so we have
$$
\frac{y}{x+a}=\frac{\frac{2 i b}{n}}{2 a}=\frac{i b}{n a}, \frac{y}{x-a}=\frac{2 b}{\frac{2 i a}{n}}=\frac{n b}{i a},
$$

Multiplying the two equations, we get the equation of the conic section $F_{2}$ that point $H$ satisfies: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
(3) According to the problem, the coordinates of point $M$ are $(0, b)$. Consider a circle with center $M$ and radius $r$, $\odot M$. The radius of $\odot M$ when it is tangent to $F_{2}$ is the minimum value we seek.
Let the equation of $\odot M$ be $x^{2}+(y-b)^{2}=r^{2}$. Then, from $\left\{\begin{array}{l}x^{2}+(y-b)^{2}=r^{2}, \\ b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2},\end{array}\right.$ we get
$$
\left(a^{2}+b^{2}\right) y^{2}-2 b^{3} y+a^{2} b^{2}+b^{4}-b^{2} r^{2}=0,
$$

So, $\Delta=4 b^{6}-4\left(a^{2}+b^{2}\right)\left(a^{2} b^{2}+b^{4}-b^{2} r^{2}\right)=0$, solving for $r^{2}=a^{2} \left\lvert\, b^{2}-\frac{b^{4}}{a^{2}+b^{2}}=\right.$ $\frac{a^{2}\left(a^{2}+2 b^{2}\right)}{a^{2}+b^{2}}$, thus, the minimum value sought is $a \sqrt{\frac{a^{2}+2 b^{2}}{a^{2}+b^{2}}}$.