If the domain of the function $f\left(x\right)=\sqrt{3{\cos }^{-1}\left(4x\right)-\pi }$  is $[a,b]$ , then the value of $(4a+64b)$  is ______

$f\left(x\right)=\sqrt{3{\cos }^{-1}\left(4x\right)-\pi }$  is defined
If  ${\cos }^{-1}4x\ge \frac{\pi }{3}\Rightarrow 4x\le \frac{1}{2}orx\le \frac{1}{8}\_\_\_\left(i\right)$
Also, $-1\le 4x\le 1or\frac{-1}{4}\le x\le \frac{1}{4}\_\_\_\_\_\left(ii\right)$
Therefore, from Eqs.(i) and(ii), we have domain: $x\in [\frac{-1}{4},\frac{1}{8}]$
 $\Rightarrow 4a+64b=7$