The equation of tangent of the circle $$x^2+y^2=4$$ is $$y=m x \pm 2 \sqrt{1+m^2}\quad \text{.... (i)}$$

Let $$y=m x \pm 2 \sqrt{1+m^2}$$ also touches $$(x-3)^2+ y^2=1$$

$$\Rightarrow(x-3)^2+\left(m x \pm 2 \sqrt{1+m^2}\right)^2=1$$

$$\Rightarrow x^2-6 x+9+m^2 x^2+4\left(1+m^2\right) \pm 4 m \sqrt{1+m^2} x=1$$

$$\Rightarrow\left(1+m^2\right) x^2+\left(-6 \pm 4 m \sqrt{1+m^2}\right) x+4\left(m^2+3\right)=0$$

Apply

$$\begin{aligned} & \quad \left(-6 \pm 4 m \sqrt{1+m^2}\right)^2-4\left(1+m^2\right) \cdot 4\left(m^2+3\right)=0 \\ \Rightarrow \quad & 36+16 m^2\left(1+m^2\right) \pm 48 m \sqrt{1+m^2} \\ & \quad -16\left(m^4+4 m^2+3\right)=0 \\ \Rightarrow \quad & 4 m^2+1= \pm 4 m \sqrt{1+m^2} \end{aligned}$$

On squaring both side

$$\begin{array}{rlrl} \Rightarrow & 16 m^4+1+8 m^2 =16 m^2+16 m^4 \\ \Rightarrow & m^2 =\frac{1}{8} \\ \Rightarrow & m = \pm \frac{1}{2 \sqrt{2}} \end{array}$$

$$\begin{array}{ll} \text { Put } & m= \pm \frac{1}{2 \sqrt{2}} \text { in the equation (i) } \\ \Rightarrow & y= \pm \frac{x}{2 \sqrt{2}} \pm \frac{6}{2 \sqrt{2}} \\ \Rightarrow & 2 \sqrt{2} y= \pm x \pm 6 \end{array}$$

Hence, the equation of common tangent of given circles are $$2 \sqrt{2} y=-x+6,2 \sqrt{2} y=x+6, 2 \sqrt{2} y=-x-6$$ and $$2 \sqrt{2} y=x-6$$.