$$\begin{gathered} \text{ If }a\text{ is an integer lying in }[-5,30],\text{ then the probability that the graph of } \\ y=x^2+2(a+4)x-5a+64\text{ is strictly above the }x\text{-axis is } \end{gathered}$$

$x^2+2(a+4)x-5a+64\ge 0$

$\text{ If }D\le 0\text{, then }$

$\begin{array}{l}(\mathrm{a}+4{)}^2-(-5\mathrm{a}+64)<0\\ \mathrm{OR} {\mathrm{a}}^2+13\mathrm{a}-48<0\\ \mathrm{OR} (\mathrm{a}+16)(\mathrm{a}-3)<0\\ \Rightarrow -16<\mathrm{a}<3\iff -5\le \mathrm{a}\le 2\end{array}$

Then, the favorable cases is equal to the number of integers in
the interval [-5, 2], i.e., 8.
Total number of cases is equal to the number of integers in
the interval [ -5, 30], i.e., 36.
Hence, the required probability is 8/36 = 2/9.