6. 4 .

Since $I F_{1}$ is the angle bisector of $\angle P F_{1} Q$, we have
$$
\frac{|P Q|}{|P I|}=1+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \text {. }
$$

Let $P\left(x_{0}, y_{0}\right)$. Then, $\left|P F_{1}\right|=\frac{3}{2} x_{0}+2$.
By the optical property of the hyperbola, the tangent line to the hyperbola at point $P$ is perpendicular to line $P Q$.
$$
\begin{array}{l}
\text { Then } l_{P Q}: y=\frac{-4 y_{0}}{5 x_{0}}\left(x-x_{0}\right)+y_{0} . \\
\text { Let } y=0 \text {, we get } Q\left(\frac{9}{4} x_{0}, 0\right) . \\
\text { Therefore, } \frac{|P Q|}{|P I|}+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}=1+2 \frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \\
=1+2 \times \frac{\frac{9}{4} x_{0}+3}{\frac{3}{2} x_{0}+2}=4 .
\end{array}
$$
6. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$, respectively. Point $P$ is on the right branch of the hyperbola $C$, and the excenter of $\triangle P F_{1} F_{2}$ opposite to $\angle P F_{1} F_{2}$ is $I$. The line $P I$ intersects the $x$-axis at point $Q$. Then
$$
\frac{|P Q|}{|P I|}+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}=
$$
$\qquad$ .