We need the quadratic $f(x)=x^2+2(a+4)x-5a+64$ to satisfy $f(x)>0 \quad \text{for all } x\in \mathbb R.$

For a quadratic $Ax^2+Bx+C$ to be positive for all real $x$, we need:

• $A>0$, and
• discriminant $<0$.

Here, $A=1>0$, so we only need $\Delta<0.$

Compute the discriminant: $\Delta=[2(a+4)]^2-4(1)(-5a+64).$

Simplifying, $\Delta=4(a+4)^2+20a-256.$

$\Delta=4(a^2+8a+16)+20a-256$ $=4a^2+32a+64+20a-256$ $=4a^2+52a-192.$

So the condition becomes $4a^2+52a-192<0.$

Divide by $4$: $a^2+13a-48<0.$

Factorize: $a^2+13a-48=(a+16)(a-3).$

Hence, $(a+16)(a-3)<0.$

Therefore, $-16<a<3.$

Now $a$ is an integer in the interval $[-5,30]$.

Integers satisfying $-16<a<3$ and $a\in[-5,30]$ are: $a=-5,-4,-3,-2,-1,0,1,2.$

Number of favorable integers $=8$.

Total integers in $[-5,30]$ are: $30-(-5)+1=36.$

Therefore, the required probability is $\frac{8}{36}=\frac{2}{9}.$

Comparing with the given options, this is Option B.