Let the equation of two diameters of a circle $$x^2 + y^2 - 2x + 2fy + 1 = 0$$ be $$2px - y = 1$$ and $$2x + py = 4p$$. Then the slope $$m \in (0, \infty)$$ of the tangent to the hyperbola $$3x^2 - y^2 = 3$$ passing through the centre of the circle is equal to ______.

The circle is $$x^2 + y^2 - 2x + 2fy + 1 = 0$$. Its center is $$(1, -f)$$ and radius is $$\sqrt{1 + f^2 - 1} = |f|$$.

Since the center $$(1, -f)$$ lies on both diameters, the equations become:

Diameter 1: $$2p(1) - (-f) = 1 \implies 2p + f = 1$$ ... (i)

Diameter 2: $$2(1) + p(-f) = 4p \implies 2 - pf = 4p$$ ... (ii)

From equation (i), we obtain $$f = 1 - 2p$$.

Substituting this into equation (ii) yields:

$$2 - p(1 - 2p) = 4p$$

$$2 - p + 2p^2 = 4p$$

$$2p^2 - 5p + 2 = 0$$

$$(2p - 1)(p - 2) = 0$$

$$p = \dfrac{1}{2} \text{ or } p = 2$$

When $$p = \dfrac{1}{2}$$, then $$f = 1 - 1 = 0$$, so the radius is 0, which is not valid.

On the other hand, if $$p = 2$$, then $$f = 1 - 4 = -3$$ and the center becomes $$(1, 3)$$.

Next, to find the tangent from the center $$(1, 3)$$ to the hyperbola, note that the hyperbola is $$3x^2 - y^2 = 3$$, or equivalently $$x^2 - \dfrac{y^2}{3} = 1$$, so $$a^2 = 1, b^2 = 3$$.

The general equation of a tangent to the hyperbola with slope $$m$$ is

$$y = mx \pm \sqrt{a^2 m^2 - b^2} = mx \pm \sqrt{m^2 - 3}$$

Since this line must pass through the point $$(1, 3)$$, we substitute to get

$$3 = m \pm \sqrt{m^2 - 3}$$

$$3 - m = \pm \sqrt{m^2 - 3}$$

Squaring both sides leads to

$$ (3 - m)^2 = m^2 - 3 $$

$$ 9 - 6m + m^2 = m^2 - 3 $$

$$ 12 = 6m $$

$$ m = 2 $$

Because $$m = 2 > 0$$ and $$\sqrt{m^2 - 3} = 1$$ satisfies $$3 - 2 = 1$$, this solution is valid.

Therefore, the answer is $$m = 2$$.