Let $$f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & x \le 1 \\ -2x + \log_2(b^2 - 4), & x > 1 \end{cases}$$. Then the set of all values of $$b$$, for which $$f(x)$$ has maximum value at $$x = 1$$, is:

We need to find all values of $$b$$ for which $$f(x)$$ has a maximum value at $$x = 1$$.

Using the first piece since $$x \le 1$$ gives $$f(1) = 1 - 1 + 10 - 7 = 3$$.

Next, for $$x \le 1$$ the function is $$f(x) = x^3 - x^2 + 10x - 7$$ and $$f'(x) = 3x^2 - 2x + 10$$. Since the discriminant of $$3x^2 - 2x + 10$$ is $$4 - 120 = -116 < 0$$ and the leading coefficient is positive, $$f'(x) > 0$$ on $$(-\infty,1]$$. This means $$f$$ is strictly increasing on $$(-\infty,1]$$, so $$f(x) \le f(1) = 3$$ for all $$x \le 1$$.

For $$x > 1$$, $$f(x) = -2x + \log_2(b^2 - 4)$$. Since the coefficient of $$x$$ is $$-2 < 0$$, this expression is strictly decreasing on $$(1,\infty)$$, so its supremum occurs as $$x \to 1^+$$: $$\lim_{x \to 1^+} f(x) = -2 + \log_2(b^2 - 4)$$.

To ensure a maximum at $$x = 1$$, we require $$f(x) \le 3$$ for all $$x > 1$$, which gives $$-2 + \log_2(b^2 - 4) \le 3$$, then $$\log_2(b^2 - 4) \le 5$$, so $$b^2 - 4 \le 32$$, hence $$b^2 \le 36$$ and $$-6 \le b \le 6$$.

Moreover, for $$\log_2(b^2 - 4)$$ to be defined we require $$b^2 - 4 > 0$$, i.e., $$b < -2$$ or $$b > 2$$. Combining this with $$-6 \le b \le 6$$ gives $$b \in [-6, -2) \cup (2, 6]$$.

The correct answer is Option C: $$[-6, -2) \cup (2, 6]$$.