The answer is $\frac{4}{3}$.
Analysis: Since the equation of circle $C$ can be transformed into $(x-4)^{2}+y^{2}=1$, the center of circle $C$ is $(4,0)$, and the radius is 1.
If there exists at least one point $A\left(x_{0}, k x_{0}-2\right)$ on the line $y=k x-2$, such that the circle with this point as the center and 1 as the radius has a common point with circle $C$, then there exists $x_{0} \in \mathbf{R}$, such that $|A C| \leq 1+1=2$ holds, i.e., $|A C|_{\min } \leq 2$.

Since $|A C|_{\min }$ is the distance $d=\frac{|4 k-2|}{\sqrt{k^{2}+1}}$ from point $C$ to the line $y=k x-2$, we have $\frac{|4 k-2|}{\sqrt{k^{2}+1}} \leq 2$. Solving this, we get $0 \leq k \leq \frac{4}{3}$, so the maximum value of $k$ is $\frac{4}{3}$.
8. Given the equation of circle $C$ as $x^{2}+y^{2}-8 x+15=0$, if there exists at least one point on the line $y=k x-2(k \in \mathbf{R})$ such that a circle with this point as its center and 1 as its radius has a common point with circle $C$, then the maximum value of $k$ is $\qquad$.