In the Cartesian coordinate system, the parabola $y=-\frac{4}{9} x^{2} +\frac{2}{9} m x+\frac{5}{9} m+\frac{4}{3}$ intersects the $x$-axis at points $A$ and $B$. It is known that point $A$ is on the negative half of the $x$-axis, and point $B$ is on the positive half of the $x$-axis, with the length of $BD$ being twice the length of $AO$ (i.e., $|BO|=2|AO|$). Point $C$ is the vertex of the parabola.
(1) Find the equation of this parabola;
(2) If point $P$ is on the axis of symmetry of this parabola and $\odot P$ is tangent to the $x$-axis and the line $BC$, find the coordinates of point $P$.


(1) Given $|O B|=2|O A|$, let $|O A|=u$. Then $|O B|=2 u (u>0)$.
Thus, $-u, 2 u$ are the roots of the equation $-\frac{4}{9} x^{2}+\frac{2}{9} m x+\frac{5}{9} m+ \frac{4}{3}=0$.
By Vieta's formulas, we have $\left\{\begin{array}{l}-u+2 u=\frac{1}{2} m, \\ -u \cdot 2 u=-\frac{5}{4} m-3 .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}m_{1}=4, \\ u_{1}=2 ;\end{array}\left\{\begin{array}{l}m_{2}=-\frac{3}{2}, \\ u_{2}=-\frac{3}{4} .\end{array}\right.\right.$ (discard the negative solution)
Thus, the equation is $y=-\frac{4}{9} x^{2}+\frac{8}{9} x+\frac{32}{9}$.
(2) Since $y=-\frac{4}{9}\left(x^{2}-2 x-8\right)$
$$
=-\frac{4}{9}(x-1)^{2}+4 \text {, }
$$

the vertex $C(1,4)$, and the axis of symmetry is $x=1$.
Let the intersection of the axis of symmetry with the $x$-axis be point $D(1,0)$.
If point $P$ is in the first quadrant, as shown in Figure 6, then we can set $P(1, r)(r>0)$. Draw $P E \perp B C$ intersecting $B C$ at $E$. By the problem statement,
$$
\begin{array}{l}
P E=P D=r, \\
D B=O B-O D \\
=4-1=3, \\
D C=4 .
\end{array}
$$

Thus, $P C=4-r, B C=\sqrt{D B^{2}+D C^{2}}=5$.
Since $\angle D C B=\angle E C P, \angle C D B=\angle C E P=90^{\circ}$, $\therefore \mathrm{Rt} \triangle B C D \sim \mathrm{Rt} \triangle P C E$.
Thus, $\frac{B D}{P E}=\frac{B C}{P C}$, i.e., $\frac{3}{r}=\frac{5}{4-r}$. $\therefore r=\frac{3}{2}$. Thus, $P\left(1, \frac{3}{2}\right)$.

If point $P$ is in the fourth quadrant, as shown in Figure 7, we can set $P(1,-r)(r>0)$.

Draw $P E \perp B C$ at point $E$, then
$$
\begin{array}{l}
P E=P D=r, \\
B E=B D=3 . \\
P C=4+r, \\
C E=C B+B E=5+3=8 .
\end{array}
$$

Similarly, we have
Rt $\triangle C D B \backsim \mathrm{Rt} \triangle C E P$.
Thus, $\frac{B D}{P E}=\frac{B C}{P C}, \frac{3}{r}=\frac{5}{4+r}, r=6$.
Hence, $P(1,-6)$.