14. Given that $F$ is the focus of the parabola $y^{2}=4 x$, point $M(4,0)$, a line with slope $k_{1}$ is drawn through point $F$ intersecting the parabola at points $A$ and $B$. Extend $A M$ and $B M$ to intersect the parabola again at points $C$ and $D$, respectively. Let the slope of line $C D$ be $k_{2}$.
(1) Find the value of $\frac{k_{1}}{k_{2}}$;
(2) Find the range of the angle $\theta$ between lines $A B$ and $C D$.
14. (1) From the given conditions, we know $F(1,0)$. Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, $C\left(x_{3}, y_{3}\right)$, and $D\left(x_{4}, y_{4}\right)$.
Without loss of generality, assume $y_{1}>0$.
By combining $l_{A B}: y=k_{1}(x-1)$ with $y^{2}=4 x$, we get $y^{2}-\frac{4}{k_{1}} y-4=0$.
Therefore, $y_{1} y_{2}=-4$, $x_{1} x_{2}=1$.
(i) If $x_{1}=4$, then $A(4,4)$.

Thus, $y_{2}=\frac{-4}{y_{1}}=-1$, $x_{2}=\frac{1}{4}$, i.e., $B\left(\frac{1}{4},-1\right)$.
From $l_{A M}: x=4$, we know $C(4,-4)$.
By combining $l_{B M}: y=\frac{4}{15}(x-4)$ with $y^{2}=4 x$, we get $y^{2}-15 y-16=0$.
Solving this, we get $y_{4}=16$, $x_{4}=64$, i.e., $D(64,16)$.
Thus, $k_{1}=\frac{4}{3}$, $k_{2}=\frac{1}{3}$.
Therefore, $\frac{k_{1}}{k_{2}}=4$.
(ii) If $x_{1} \neq 4$, by combining $l_{A M}: y=\frac{y_{1}}{x_{1}-4}(x-4)$ with $y^{2}=4 x$, we get
$$
y_{1}^{2}(x-4)^{2}=4 x\left(x_{1}-4\right)^{2} \text {. }
$$

Since $y_{1}^{2}=4 x_{1}$, simplifying gives
$$
x_{1} x^{2}-\left(x_{1}^{2}+16\right) x+16 x_{1}=0 \text {. }
$$

This equation has one root $x_{1}$, so the other root is
$$
x_{3}=\frac{16}{x_{1}}, y_{3}=\frac{-16}{y_{1}} \text {. }
$$

Thus, $C\left(\frac{16}{x_{1}},-\frac{16}{y_{1}}\right)$.
Similarly, $D\left(\frac{16}{x_{2}},-\frac{16}{y_{2}}\right)$.
$$
=\frac{-\frac{16}{y_{2}}+\frac{16}{y_{1}}}{\frac{16}{x_{2}}-\frac{16}{x_{1}}}=-\frac{x_{1} x_{2}}{y_{1} y_{2}} \cdot \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1}{4} k_{1} \Rightarrow \frac{k_{1}}{k_{2}}=4.
$$
From (i) and (ii), we know $\frac{k_{1}}{k_{2}}=4$.
(2) Note that
$\tan \theta=\left|\frac{k_{1}-k_{2}}{1+k_{1} k_{2}}\right|=\left|\frac{3 k_{1}}{4+k_{1}^{2}}\right| \leqslant \frac{3}{4}$.
Thus, $\theta \leqslant \arctan \frac{3}{4}$.
Therefore, the range of the angle $\theta$ is $\left(0, \arctan \frac{3}{4}\right]$.