Step 1: Apply the domain rule for arccos.
$\cos^{-1}(u)$ needs $-1 \le u \le 1$. Here $u = \log_2(x^2+5x+8)$, so we require $-1 \le \log_2(x^2+5x+8) \le 1$.
Step 2: Remove the logarithm.
Since base $2 > 1$, exponentiate keeping the direction: $2^{-1} \le x^2+5x+8 \le 2^{1}$, i.e. $\tfrac12 \le x^2+5x+8 \le 2$.
Step 3: Handle the upper bound.
$x^2+5x+8 \le 2 \Rightarrow x^2+5x+6 \le 0 \Rightarrow (x+2)(x+3) \le 0$, giving $-3 \le x \le -2$.
Step 4: Handle the lower bound.
$x^2+5x+8 \ge \tfrac12 \Rightarrow x^2+5x+7.5 \ge 0$. Its discriminant is $25 - 30 = -5 < 0$, and the parabola opens upward, so this is true for every real $x$.
Step 5: Intersect the conditions.
The lower-bound condition is always satisfied, so the domain is just the upper-bound result $[-3, -2]$.
Step 6: State the domain.
The domain is $[-3, -2]$, which is option (B).
\[ \boxed{[-3,\ -2]} \]