2.2 In the Cartesian coordinate system $x O y$, given three points $A\left(0, \frac{4}{3}\right), B(-1,0)$, $C(1,0)$. The distance from point $P$ to line $B C$ is the geometric mean of the distances from the point to lines $A B$ and $A C$.
(1) Find the equation of the locus of point $P$;
(2) If line $l$ passes through the incenter of $\triangle A B C$ (denoted as $D$), and intersects the locus of point $P$ at exactly 3 points, find the range of the slope $k$ of $l$.
(1) The equations of the lines $AB$, $AC$, and $BC$ are $y=\frac{4}{3}(x+1)$, $y=-\frac{4}{3}(x-1)$, and $y=0$, respectively. The distances from point $P(x, y)$ to $AB$, $AC$, and $BC$ are $d_{1}=\frac{1}{5}|4 x-3 y+4|$, $d_{2}=\frac{1}{5}|4 x+3 y-4|$, and $d_{3}=|y|$, respectively.

According to the problem, $d_{1} d_{2}=d_{3}^{2}$, which gives $\left|16 x^{2}-(3 y-4)^{2}\right|=25 y^{2}$. This can be simplified to $16 x^{2}-(3 y-4)^{2}+25 y^{2}=0$ or $16 x^{2}-(3 y-4)^{2}-25 y^{2}=0$. Simplifying further, the trajectory of point $P$ is given by the circle $S: 2 x^{2}+2 y^{2}+3 y-2=0$ and the hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$.
(2) From the above, the trajectory of point $P$ consists of two parts: the circle $S: 2 x^{2}+2 y^{2}+3 y-2=0$,
(1) and the hyperbola $T: 8 x^{2}-17 y^{2}+12 y-8=0$. (2) Since points $B(-1,0)$ and $C(1,0)$ satisfy the given conditions, points $B$ and $C$ lie on the trajectory of point $P$, and the curves $S$ and $T$ intersect only at points $B$ and $C$. The incenter $D$ of $\triangle ABC$ also satisfies the given conditions, and from $d_{1}=d_{2}=d_{3}$, we find $D\left(0, \frac{1}{2}\right)$, which lies on the circle $S$. The line $l$ passes through $D$ and intersects the trajectory of point $P$ at 3 points, so the slope of $l$ exists. Let the equation of $l$ be $y=k x+\frac{1}{2}$.
(i) When $k=0$, $l$ is tangent to the circle $S$ at a single point $D$; in this case, the line $y=\frac{1}{2}$ is parallel to the $x$-axis, indicating that $l$ intersects the hyperbola $T$ at two points different from $D$, so $l$ intersects the trajectory of point $P$ at exactly 3 points.
(ii) When $k \neq 0$, $l$ intersects the circle $S$ at two different points. In this case, $l$ intersects the trajectory of point $P$ at exactly 3 points in two scenarios:

Scenario 1: The line $l$ passes through point $B$ or point $C$, in which case the slope $k= \pm \frac{1}{2}$. The equation of line $l$ is $x=\pm(2 y-1)$. Substituting into equation (2) gives $y(3 y-4)=0$, solving for $E\left(\frac{5}{3}, \frac{4}{3}\right)$ or $F\left(-\frac{5}{3}, \frac{4}{3}\right)$. This indicates that line $BD$ intersects the curve $T$ at 2 points $B$ and $E$, and line $CD$ intersects the curve $T$ at 2 points $C$ and $F$. Therefore, when $k= \pm \frac{1}{2}$, $l$ intersects the trajectory of point $P$ at exactly 3 points.

Scenario 2: The line $l$ does not pass through points $B$ and $C$ (i.e., $k \neq \pm \frac{1}{2}$). Since $l$ intersects $S$ at two different points, $l$ must intersect the hyperbola $T$ at exactly one point. That is, the system of equations
\[
\left\{
\begin{array}{l}
8 x^{2}-17 y^{2}+12 y-8=0, \\
y=k x+\frac{1}{2}
\end{array}
\right.
\]
has a unique real solution if and only if $8-17 k^{2}=0$ (4) or $\left(12 k^{2}-8\right)^{2}-4(8-17 k^{2})(2-8 k^{2})=0$ (5). Solving equation (4) gives $k= \pm \frac{2 \sqrt{34}}{17}$, and solving equation (5) gives $k= \pm \frac{\sqrt{2}}{2}$. In summary, the set of possible values for the slope $k$ of line $l$ is the finite set $\left\{0, \pm \frac{1}{2}, \pm \frac{2 \sqrt{34}}{17}, \pm \frac{\sqrt{2}}{2}\right\}$.