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