15. Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its left and right foci as $F_{1}$ and $F_{2}$, respectively, and its right vertex as $A$. Let $P$ be any point on the ellipse $C$. It is known that the maximum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is 3, and the minimum value is 2.
(1) Find the equation of the ellipse $C$;
(2) If the line $l: y=k x+m$ intersects the ellipse $C$ at points $M$ and $N$ (where $M$ and $N$ are not the left or right vertices), and the circle with diameter $MN$ passes through point $A$, prove that the line $l$ passes through a fixed point, and find the coordinates of this fixed point.
15. (1) Since $P$ is any point on the ellipse, we have
$$
\begin{array}{l}
\left|P F_{1}\right|+\left|P F_{2}\right|=2 a, \\
a-c \leqslant\left|P F_{1}\right| \leqslant a+c .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=\left|\overrightarrow{P F_{1}}\right|\left|\overrightarrow{P F_{2}}\right| \cos \angle F_{1} P F_{2} \\
=\frac{1}{2}\left(\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-4 c^{2}\right) \\
=\frac{1}{2}\left(\left|P F_{1}\right|^{2}+\left(2 a-\left|P F_{1}\right|\right)^{2}-4 c^{2}\right) \\
=\left(\left|P F_{1}\right|-a\right)^{2}+a^{2}-2 c^{2} .
\end{array}
$$

When $\left|P F_{1}\right|=a$, $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ reaches its minimum value $a^{2}-2 c^{2}=3$;

When $\left|P F_{1}\right|=a-c$ or $a+c$, $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ reaches its maximum value $a^{2}-c^{2}=2$.
$$
\begin{array}{l}
\text { Hence }\left\{\begin{array} { l } 
{ a ^ { 2 } - c ^ { 2 } = 3 , } \\
{ a ^ { 2 } - 2 c ^ { 2 } = 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2}=4, \\
c^{2}=1
\end{array}\right.\right. \\
\Rightarrow b^{2}=a^{2}-c^{2}=3 .
\end{array}
$$

Therefore, the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.

Substituting $y=k x+m$ into the ellipse equation, we get
$$
\begin{array}{l}
\left(4 k^{2}+3\right) x^{2}+8 k m x+4 m^{2}-12=0 \\
\Rightarrow x_{1}+x_{2}=\frac{-8 k m}{4 k^{2}+3}, x_{1} x_{2}=\frac{4 m^{2}-12}{4 k^{2}+3} .
\end{array}
$$

From $y_{1}=k x_{1}+m, y_{2}=k x_{2}+m$, we know
$$
y_{1} y_{2}=k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2} \text {. }
$$

Since the circle with diameter $M N$ passes through point $A$, we have
$$
\begin{array}{l}
\overrightarrow{A M} \cdot \overrightarrow{A N}=0 \\
\Rightarrow 7 m^{2}+16 k m+4 k^{2}=0
\end{array}
$$
$\Rightarrow m=-\frac{2}{7} k$ or $m=-2 k$ both satisfy $\Delta>0$.
If $m=-2 k$, the line $l$ always passes through the fixed point $(2,0)$, which is not acceptable;

If $m=-\frac{2}{7} k$, the line $l: y=k\left(x-\frac{2}{7}\right)$ always passes through the fixed point $\left(\frac{2}{7}, 0\right)$.