The function $$f(x) = x^3 - 6x^2 + ax + b$$ is such that $$f(2) = f(4) = 0$$. Consider two statements:
$$S_1$$: there exists $$x_1, x_2 \in (2, 4)$$, $$x_1 \lt x_2$$, such that $$f'(x_1) = -1$$ and $$f'(x_2) = 0$$.
$$S_2$$: there exists $$x_3, x_4 \in (2, 4)$$, $$x_3 \lt x_4$$, such that $$f$$ is decreasing in $$(2, x_4)$$, increasing in $$(x_4, 4)$$ and $$2f'(x_3) = \sqrt{3}f(x_4)$$. Then

We are given the function $$f(x) = x^3 - 6x^2 + ax + b$$ such that $$f(2) = f(4) = 0$$.

Substituting $$x = 2$$ into the function: $$f(2) = 2^3 - 6(2^2) + a(2) + b = 8 - 24 + 2a + b = 2a + b - 16 = 0$$, which gives the equation $$2a + b = 16$$.

Substituting $$x = 4$$ into the function: $$f(4) = 4^3 - 6(4^2) + a(4) + b = 64 - 96 + 4a + b = 4a + b - 32 = 0$$, which gives the equation $$4a + b = 32$$.

Solving these two linear equations simultaneously, we subtract the first from the second: $$(4a + b) - (2a + b) = 32 - 16$$, which results in $$2a = 16$$ or $$a = 8$$.

Substituting $$a = 8$$ into $$2a + b = 16$$ gives $$16 + b = 16$$, so $$b = 0$$.

Thus, the function is $$f(x) = x^3 - 6x^2 + 8x$$.

Differentiating $$f(x)$$ with respect to $$x$$, we get the derivative function: $$f'(x) = 3x^2 - 12x + 8$$.

Now let us analyze statement $$S_1$$: "there exists $$x_1, x_2 \in (2, 4), x_1 < x_2$$, such that $$f'(x_1) = -1$$ and $$f'(x_2) = 0$$."

At the endpoints, $$f(2) = 0$$ and $$f(4) = 0$$. By Rolle's Theorem, since $$f(x)$$ is continuous on $$[2, 4]$$ and differentiable on $$(2, 4)$$, there exists a value $$c \in (2, 4)$$ such that $$f'(c) = 0$$.

Solving $$3x^2 - 12x + 8 = 0$$, we find the roots using the quadratic formula: $$x = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2}{\sqrt{3}}$$.

Taking $$x_2 = 2 + \frac{2}{\sqrt{3}} \approx 3.15$$, this value lies in the interval $$(2, 4)$$ and satisfies $$f'(x_2) = 0$$.

To check for $$f'(x_1) = -1$$, we evaluate the derivative at the lower endpoint: $$f'(2) = 3(4) - 12(2) + 8 = 12 - 24 + 8 = -4$$.

Since $$f'(2) = -4$$ and $$f'(x_2) = 0$$, by the Intermediate Value Theorem for continuous derivatives, there exists $$x_1 \in (2, x_2)$$ such that $$f'(x_1) = -1$$. Therefore, statement $$S_1$$ is true.

Now let us analyze statement $$S_2$$: "there exists $$x_3, x_4 \in (2, 4), x_3 < x_4$$, such that $$f$$ is decreasing in $$(2, x_4)$$, increasing in $$(x_4, 4)$$ and $$2f'(x_3) = \sqrt{3}f(x_4)$$."

For monotonicity, we examine the sign of $$f'(x) = 3x^2 - 12x + 8$$. The critical point in $$(2, 4)$$ is $$x_4 = 2 + \frac{2}{\sqrt{3}}$$. Since $$f'(2) = -4 < 0$$ and the parabola opens upwards, $$f(x)$$ is decreasing on $$(2, x_4)$$ and increasing on $$(x_4, 4)$$.

Let us compute $$f(x_4)$$ at $$x_4 = 2 + \frac{2}{\sqrt{3}}$$: $$f(x_4) = x_4(x_4 - 2)(x_4 - 4) = \left(2 + \frac{2}{\sqrt{3}}\right) \left(\frac{2}{\sqrt{3}}\right) \left(\frac{2}{\sqrt{3}} - 2\right) = \frac{-16}{3\sqrt{3}}$$.

Thus, the right side of the condition is $$\sqrt{3}f(x_4) = \sqrt{3}\left(-\frac{16}{3\sqrt{3}}\right) = -\frac{16}{3}$$.

We need $$2f'(x_3) = -\frac{16}{3}$$, which means $$f'(x_3) = -\frac{8}{3} \approx -2.67$$.

Since $$f'(2) = -4$$ and $$f'(x_4) = 0$$, and $$-\frac{8}{3}$$ lies between $$-4$$ and $$0$$, by the Intermediate Value Theorem, there exists $$x_3 \in (2, x_4)$$ such that $$f'(x_3) = -\frac{8}{3}$$.

Since $$x_3 \in (2, x_4)$$ and $$x_4 < 4$$, we have $$x_3 < x_4$$. Therefore, statement $$S_2$$ is true.

Hence both statements $$S_1$$ and $$S_2$$ are true, making the correct option C.