The probability of selecting integers $a\in \left[-5,30\right]$ such that $x^2+2\left(a+4\right)x-5a+64>0$ for all $x\in R$

$a\in \left[-5,30\right]$

$S=\left\{-5,-4,-3,---0,1,2,---30\right\}$

$n\left(s\right)=36$

E is the event, satisfying

$x^2+2\left(a+4\right)x-59+64>0$

$\Delta <0\Rightarrow 4{\left(a+4\right)}^2-4\left(1\right)\left(64-5a\right)<0$

$\Rightarrow a^2+13a-48<0$

$\left(a+16\right)\left(a-3\right)<0$

$a\in \left(-16,3\right)$

But $a\in \left[-5,30\right]$

$\therefore a\in \left[-5,3\right]$

$\therefore E=\left\{-5,-4,-3,-2,-1,0,1,2\right\}$

$n\left(E\right)=8$

$\therefore P\left(E\right)=\frac{8}{36}=\frac{2}{9}$