Given that $$(g \circ f)(x) = x$$ for all $$x \in \mathbf{R}$$. This means $$g(f(x)) = x$$ for all $$x \in \mathbf{R}$$. Differentiating both sides with respect to $$x$$, we get:

$$g'(f(x)) \cdot f'(x) = 1$$

Now, we want to find the value of $$8g'(2)$$. To do this, we need to find a value of $$x$$ such that $$f(x) = 2$$. Let's solve for $$x$$:

$$x^5 + 2e^{x/4} = 2$$

By inspection, we see that $$x = 0$$ is a solution. Therefore, $$f(0) = 2$$. Now, we can substitute this into our differentiated equation:

$$g'(f(0)) \cdot f'(0) = 1$$

$$g'(2) \cdot f'(0) = 1$$

Let's find $$f'(0)$$:

$$f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$$

$$f'(0) = \frac{1}{2}$$

Substituting this back into our equation:

$$g'(2) \cdot \frac{1}{2} = 1$$

$$g'(2) = 2$$

Finally, we can calculate $$8g'(2)$$:

$$8g'(2) = 8 \cdot 2 = \boxed{16}$$