Determine the domain for the first term, g(x) = ^-1((2-|x|)/(4)).

For $\cos^{-1}(u)$ to be defined, the argument $u$ must satisfy $-1 \le u \le 1$.
So, we must have:

$$-1 \le \frac{2-|x|}{4} \le 1$$

Multiply all parts by 4:

$$-4 \le 2-|x| \le 4$$

Subtract 2 from all parts:

$$-4-2 \le -|x| \le 4-2$$ $$-6 \le -|x| \le 2$$

Multiply all parts by -1 and reverse the inequality signs:

$$-2 \le |x| \le 6$$

Since $|x|$ is always non-negative, the condition $|x| \ge -2$ is always true. Thus, we only need to consider $|x| \le 6$.
This implies $-6 \le x \le 6$.
So, the domain for the first term is $D_1 = [-6, 6]$.

For this term to be defined, two conditions must be met:

1. The argument of the logarithm must be positive: $$3-x > 0 \implies x < 3$$
2. The denominator cannot be zero, meaning $\log_e(3-x) \ne 0$. This implies $3-x \ne e^0$. $$3-x \ne 1$$ $$x \ne 2$$

So, the domain for the second term is $D_2 = (-\infty, 3) \setminus \{2\}$.

The domain of $f(x)$ is the intersection of $D_1$ and $D_2$:

$$D = D_1 \cap D_2 = [-6, 6] \cap ((-\infty, 3) \setminus \{2\})$$

The intersection of $[-6, 6]$ and $(-\infty, 3)$ is $[-6, 3)$.
Then, we must exclude $x=2$ from this interval.
So, the domain is $D = [-6, 3) \setminus \{2\}$, which can be written as $[-6, 3] - \{2\}$.

By comparing $D = [-6, 3] - \{2\}$ with $[-\alpha, \beta] - \{\gamma\}$, we can identify the values:

$$-\alpha = -6 \implies \alpha = 6$$ $$\beta = 3$$ $$\gamma = 2$$

$$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$$

The final answer is $\boxed{\text{11}}$.