13. Given the hyperbola \( C: \frac{x^{2}}{4}-y^{2}=1 \), a line \( l \) passing through the point \( M(1,-1) \) intersects the right branch of the hyperbola \( C \) at points \( A \) and \( B \), and intersects the \( x \)-axis at point \( N \). Let
\[
\overrightarrow{M A}=\lambda_{1} \overrightarrow{A N}, \overrightarrow{M B}=\lambda_{2} \overrightarrow{B N}\left(\lambda_{1}, \lambda_{2} \in \mathbf{R}\right) \text {. }
\]

Find the range of \(\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}\).
13. It is known that the slope of the line exists, let it be $k$.

Then the line $l: y+1=k(x-1), N\left(\frac{1}{k}+1,0\right)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Substituting the line $l$ into the equation of the hyperbola $C$ yields
$\left(1-4 k^{2}\right) x^{2}+8 k(k+1) x-$
$4(k+1)^{2}-4=0$.
From the equation (1) having two distinct positive roots $x_{1} 、 x_{2}$, we know
$\left\{\begin{array}{l}\Delta>0, \\ x_{1}+x_{2}=\frac{-8 k(k+1)}{1-4 k^{2}}>0, \\ x_{1} x_{2}=\frac{-4(k+1)^{2}-4}{1-4 k^{2}}>0\end{array}\right.$
$\Rightarrow \frac{1}{2}<k<\frac{1+\sqrt{7}}{3}$.
Let $t=\frac{1}{k}+1$. Then $t \in\left(\frac{\sqrt{7}+1}{2}, 3\right)$.
By $\overrightarrow{M A}=\lambda_{1} \overrightarrow{A N} \Rightarrow x_{1}=\frac{1+\lambda_{1} t}{1+\lambda_{1}}, y_{1}=\frac{-1}{1+\lambda_{1}}$.
Substituting into the equation of the hyperbola $C$, and simplifying, we get
$$
\left(t^{2}-4\right) \lambda_{1}^{2}+(2 t-8) \lambda_{1}-7=0 \text {. }
$$

Similarly, $\left(t^{2}-4\right) \lambda_{2}^{2}+(2 t-8) \lambda_{2}-7=0$.
Thus, $\lambda_{1} 、 \lambda_{2}$ are the two real roots of the quadratic equation
$$
\left(t^{2}-4\right) \lambda^{2}+(2 t-8) \lambda-7=0
$$

By Vieta's formulas, we have
$$
\begin{array}{l}
\lambda_{1}+\lambda_{2}=\frac{2 t-8}{4-t^{2}}, \lambda_{1} \lambda_{2}=\frac{7}{4-t^{2}} \\
\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(\lambda_{1}+\lambda_{2}\right)^{2}}{\lambda_{1} \lambda_{2}}-2=\frac{(2 t-8)^{2}}{7\left(4-t^{2}\right)}-2 .
\end{array}
$$

Let $z=t-4$. Then
$$
\begin{array}{l}
z \in\left(\frac{\sqrt{7}-7}{2},-1\right), \\
\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}=-\frac{4}{7\left(\frac{12}{z^{2}}+\frac{8}{z}+1\right)}-2 .
\end{array}
$$

Let $w=\frac{1}{z}$. Then $w \in\left(-1,-\frac{\sqrt{7}+7}{21}\right)$.
Let $f(w)=12 w^{2}+8 w+1$, whose axis of symmetry is $w=-\frac{1}{3}$.
Clearly, the range of $w$ is to the left of $w=-\frac{1}{3}$, thus, $f\left(-\frac{\sqrt{7}+7}{21}\right)<f(w)<f(-1)$, i.e.,
$$
\begin{array}{l}
-\frac{1}{7}<f(w)<5, \text { and } f(w) \neq 0 . \\
\text { Hence } \frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}=-\frac{4}{7 f(w)}-2 \\
\in\left(-\infty,-\frac{74}{35}\right) \cup(2,+\infty) .
\end{array}
$$