21. (12 points) Given points $F_{1}(-1,0), F_{2}(1,0)$,
$\odot F_{2}:(x-1)^{2}+y^{2}=1$,
a moving circle on the right side of the $y$-axis is tangent to the $y$-axis and externally tangent to $\odot F_{2}$. The locus of the center of this moving circle is curve $C$, and curve $E$ is an ellipse with foci at $F_{1}$ and $F_{2}$.
(1) Find the equation of curve $C$;
(2) Let the intersection point of curve $C$ and curve $E$ in the first quadrant be $P$, and $\left|P F_{1}\right|=\frac{7}{3}$. Find the standard equation of curve $E$;
(3) Under the conditions of (1) and (2), a line $l$ intersects the ellipse $E$ at points $A$ and $B$. If the midpoint $M$ of $A B$ lies on curve $C$, find the range of the slope $k$ of line $l$.
21. (1) Let the center of the moving circle be $D(x, y) (x>0)$.

Since $\odot D$ is tangent to the $y$-axis on its right side and is externally tangent to $\odot F_{2}$, we have $\left|D F_{2}\right|-x=1$.
Thus, $\sqrt{(x-1)^{2}+y^{2}}=x+1$.
Therefore, the equation of curve $C$ is $y^{2}=4 x (x>0)$.
(2) Since curve $E$ is an ellipse, we know $c=1$.
Given $\left|P F_{1}\right|=\frac{7}{3}$, we have $x_{P}=\frac{2}{3}$.
Thus, $\left|P F_{2}\right|=\frac{5}{3}$.
By the definition of an ellipse,
$2 a=\left|P F_{1}\right|+\left|P F_{2}\right|=\frac{7}{3}+\frac{5}{3}=4 \Rightarrow a=2$.
Hence, $b^{2}=a^{2}-c^{2}=3$.
Therefore, the standard equation of curve $E$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(3) Let the intersection points of line $l$ with ellipse $E$ be $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and the midpoint of $A B$ be $M\left(x_{0}, y_{0}\right)$.
Assume the equation of line $l$ is
$y=k x+m (k \neq 0, m \neq 0)$.
Combining with $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, we get
$\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-12=0$.
From $\Delta>0$, we have
$4 k^{2}-m^{2}+3>0$.
By the relationship between roots and coefficients, we know
$x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$.
Thus, $x_{0}=-\frac{4 k m}{3+4 k^{2}}, y_{0}=\frac{3 m}{3+4 k^{2}}$.
Substituting point $M\left(-\frac{4 k m}{3+4 k^{2}}, \frac{3 m}{3+4 k^{2}}\right)$ into $y^{2}=4 x$ and simplifying, we get
$m=-\frac{16 k\left(3+4 k^{2}\right)}{9}$.
Substituting equation (2) into equation (1), we get
$$
\begin{array}{l}
16^{2} k^{2}\left(3+4 k^{2}\right)0) \text {. Then } \\
64 t^{2}+192 t-81<0 \\
\Rightarrow 0<t<\frac{3}{8} .
\end{array}
$$

Thus, $-\frac{\sqrt{6}}{8}<k<\frac{\sqrt{6}}{8}$, and $k \neq 0$.