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

$$\begin{gathered} \in \left[-5, 30\right] \\ \mathrm{n}\left(\mathrm{S}\right)=36 \end{gathered}$$

E=Set of values of x satisfy 

$$\begin{gathered} {\mathrm{x}}^2+2(\mathrm{a}+4)\mathrm{x}-5\mathrm{a}+64>0 \\ ∆<0 \Rightarrow 4{(\mathrm{a}+4)}^2-4(-5\mathrm{a}+64)<0 \\ \Rightarrow {\mathrm{a}}^2+16+8\mathrm{a}+5\mathrm{a}-64<0 \\ \Rightarrow {\mathrm{a}}^2+13\mathrm{a}-48<0 \\ \Rightarrow (\mathrm{a}+16) (\mathrm{a}-3)<0 \\ \Rightarrow -16<\mathrm{a}<3 \\ \mathrm{a}= \{ -5, -4, -3, -2, -1, 0, 1, 2 \} \\ \mathrm{n}\left(\mathrm{E}\right)=8 \\ \mathrm{P}\left(\mathrm{E}\right)=\frac{8}{36}=\frac{2}{9} \end{gathered}$$