Removing Discontinuities Practice Test

For $t(x)=\begin{cases}\dfrac{x^2-64}{x-8},&x\ne8\\0,&x=8\end{cases}$, what value should replace $0$ to make $t$ continuous at $x=8$?

Let $f(x)=\begin{cases}\dfrac{x^2-2x-15}{x-5},&x\ne5\\12,&x=5\end{cases}$. What value should replace $12$ to remove the discontinuity at $x=5$?