Testing the three-part continuity definition at x = 3

First check part (1): $h(3)$ is given directly as $1$, so this part is satisfied. Next examine the one-sided limits. For $x>3$, $x-3>0$, so $|x-3|=x-3$ and $h(x) = \frac{x-3}{x-3}=1$, giving $\lim_{x\to 3^+} h(x) = 1$. For $x<3$, $x-3<0$, so $|x-3|=-(x-3)$ and $h(x) = \frac{-(x-3)}{x-3} = -1$, giving $\lim_{x\to 3^-} h(x) = -1$.

Since the left-hand limit ($-1$) and right-hand limit ($1$) are different finite values, $\lim_{x\to 3} h(x)$ does not exist, so part (2) of the continuity definition fails. Because both one-sided limits exist but are unequal, $h$ has a jump discontinuity at $x=3$; it is not continuous there despite $h(3)$ being defined.

No. The discontinuity is a jump, not removable, because the left-hand and right-hand limits are unequal finite numbers. Redefining h of three would satisfy part one, but part two still fails since the overall limit at three still would not exist. Only removable discontinuities can be fixed by redefining a single point's value.