Vertical Asymptotes and Holes

There is a hole at $x=2$ because the factor cancels completely.

There is both a vertical asymptote and a hole at $x=2$.

The function is defined at $x=2$, so it has neither a hole nor a vertical asymptote.

There is a vertical asymptote at $x=2$ because the factor $x-2$ cancels partially (the numerator cancels one instance) leaving an uncanceled factor in the denominator.