Solutions

i)

$$f(x)=\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$$

$f$ is already continuous on $\mathbb{R}$. Hence, the continuous function $F$ we're looking for is just $f$.

ii) There is no such function $F$ because

$$\lim_{x \to 0^-} f(x)=-1 \ne 1=\lim_{x \to 0^+} f(x).$$

iii) Let $F(x)=0$.

iv) No such function $F$.