1. We need the domain of $f(x)=4^{-x^2}+\cos^{-1}\left(\frac{x}{2}-1\right)+\log(\cos x)$ restricted to the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$

2. Check each term separately.

(i) Term $4^{-x^2}$

This is an exponential expression and is defined for every real $x$. So, no restriction from this term.

(ii) Term $\cos^{-1}\left(\frac{x}{2}-1\right)$

For $\cos^{-1}(t)$ to be defined, we need $-1 \le t \le 1.$ Here, $t=\frac{x}{2}-1.$ So, $-1 \le \frac{x}{2}-1 \le 1.$ Add $1$ throughout: $0 \le \frac{x}{2} \le 2.$ Multiply by $2$: $0 \le x \le 4.$ Thus this term requires $x\in[0,4].$

(iii) Term $\log(\cos x)$

For logarithm to be defined, $\cos x>0.$ Now on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$ we know $\cos x>0$ for all $x$ in this open interval. So this term requires $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$

3. Combine all restrictions:

• from exponential: all real numbers,
• from inverse cosine: $[0,4]$,
• from logarithm and given interval: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Hence the required interval is $[0,4]\cap \left(-\frac{\pi}{2},\frac{\pi}{2}\right)=[0,\tfrac{\pi}{2}).$

4. Match with options: $\boxed{\left[0,\frac{\pi}{2}\right)}$ which is Option B.