5. As shown in Figure $1, \odot O_{1}$ is externally tangent to $\odot O_{2}$ at point $P$. From point $A$ on $\odot O_{1}$, a tangent line $A B$ is drawn to $\odot O_{2}$, with $B$ as the point of tangency. Line $A P$ is extended to intersect $\odot O_{2}$ at point $C$. Given that the radii of $\odot O_{1}$ and $\odot O_{2}$ are $2$ and $1$ respectively, then
$$
\frac{A C}{A B}=
$$
$\qquad$
5. $\frac{\sqrt{6}}{2}$.

Connect $O_{1} O_{2}$. Then $O_{1} O_{2}$ passes through the point of tangency $P$.
Connect $O_{1} A$, $O_{1} P$, $O_{2} P$, $O_{2} C$.
Then isosceles $\triangle O_{1} A P \backsim$ isosceles $\triangle O_{2} P C$
$$
\Rightarrow \frac{A P}{P C}=\frac{O_{1} P}{O_{2} P}=2 \text {. }
$$

Assume $A P=2 t$. Then $P C=t$.
By the secant-tangent theorem,
$$
\begin{array}{l}
A B^{2}=A P \cdot A C=6 t^{2} \Rightarrow A B=\sqrt{6} t . \\
\text { Therefore, } \frac{A C}{A B}=\frac{3 t}{\sqrt{6} t}=\frac{\sqrt{6}}{2} .
\end{array}
$$