(a) "Order preserving" implies "injective". Therefore, $f$ is bijective. Order preserving also implies that $f$ preserves the order topology.(b) It is increasing from $0$ to $+\infty$ (this part may require using the intermediate value theorem if it was not proved before), therefore, it is order preserving and surjective and, by (a), homeomorphism. Therefore, its inverse is continuous.(c) $f$ is obviously increasing and surjective. In the order topology on $X$ $[0,+\infty)$ would not be open, while it is open in the subspace topology. Therefore, $f$ would be a homeomorphism had $X$ the order topology, but in the subspace topology there are too many open sets in $X$ which makes $f^{-1}$ not continuous: $[0,+\infty)$ is open in the subspace topology (not in the order topology) but $f([0,+\infty))=[0,+\infty)$ which is not open in $\mathbb{R}$ .