Common people, this is a good problem. Think on it !!

Actually, 1) implies again 2). Here is a proof: By composing with a translation one can assume that $f$ has a fixed point $O$. Put $g(v)=f(O+v)$. We will prove that $g$ is linear by induction. Let us do first for $n=2$. Because $f$ is injective, two parallel lines are taken into two parallel lines. This shows that $g(rv)$ is collinear to $g(v)$ $g(rv)=h(r)g(v)$ for some real $h(r)$. Now, take $u$ not collinear to $v$ . Using Thales it follows that $g(ru)=h(r)g(u)$. It is immediate to prove that this result holds for $u$ collinear to $v$ (pass by some other vector which is not collinear to any of them). Next, using this property, it is clear that $h$ is multiplicative. Now, take $u,v$ not collinear. By looking at the parallelogram $OACB$ with $A=O+u$, $B=O+v$, $C=O+u+v$ it follows that $g(u)+g(v)=g(u+v)$. It is not difficult (simple algebraic manipulation) to prove that the result holds for collinear $u,v$: take $w$ not collinear to them and write $g(u+v)=g(u+(v/2+w)+(v/2-w))=g(u+(v/2+w))+g(v/2-w)=...=g(u)+g(v)$. Thus $h$ is additive and multiplicative thus it is the identity. Thus $g$ is linear and conclusion for $n=2$. Now, pass to general case. Take any hyperplan $H$, then $g$ restricted to $H$ is linear by induction and by composing with a linear map we may assume that $f(H)=H$. By considering a hyperplan that contains $u,v$ and using this observation, we deduce that $g(au+bv)=ag(u)+bg(v)$. This shows that $g$ is linear and finishes the proof. Unfortunately, for a) my solution is quite cumbersome and I prefer to see some easier proof.

by induction ? the image of any $n-2$ dimensional subspace is a $n-2$ dimensional subspace: take $U$ such a $n-2$ dimensional subspace, $U=H_{1}\cap H_{2}$ where $H_{i}$ are hyperplanes. Because $f$ is a bijection, $f(U) = f(H_{1}) \cap f(H_{2})$, where $f(H_{i})$ are distinct hyperplanes. Continue like this and I think the result will follow (I've not checked)