For part i, I can use the straight line homotopy from $f(x)$ to the antipodal, and then normalise it, then because there are no fixed points in $f$ the denominator of this will not equal 0 so its well defined.
I am pretty stuck on the next parts though.
Let us generalize the construction above by proving the following lemma
We prove this the same way as you did earlier, by using a straightline homotopy and normalizing. Explicitly this is given by $$x \mapsto \frac{(1-t)f(x) + tg(x)}{||(1-t)f(x) + tg(x)||}$$ Note that this works because $f(x)$ is never equal to $-g(x)$.
If we let $f,g$ be an aribtrary $f$ and the antipodal map (which we will denote $r$), then using this lemma shows that $f \sim r$ if $f$ has no fixed point, which is what you did.
Now let us use the lemma with $f$ an arbitrary $f$ and $g = id_{S^n}$. Then the lemma reads that $f \sim g$ if $f(x) \neq -id_{S^n}(x)$ i.e. $f(x) \neq r(x)$ for all x. In other words, if $f$ has no antipodes, then $f$ is homotopic to the identity.
So, let us assume that we have a map $f$ that has no fixed point. So $f$ is homotopic to the antipodal map. But $f$ also has no antipodes so $f$ is homotopic to the identity. We are told that these two maps are not homotopic for even $n$, so this is a contradiction. So $f$ has to have either a fixed point or an antipode.
Lastly, let us exhibit such a map for odd $n$. Consider $S^n \subset \mathbb{R}^{n+1}$. If $n = 2k+1$ then the map $(x_0, x_1,...,x_{2k+1}) \mapsto (x_1,-x_0,...,x_{2k+1},-x_{2k})$ is a map with no fixed points and no antipodes.
The justification about why the above map has no antipodes or fixed points is a little geometric. If you let $v(x)$ denote the above map, then $x \cdot v(x) = 0$ so $v(x)$ is always perpendicular to $x$. However, $x$ and $-x$ are parallel to $x$, so we know that $v$ has no fixed points or antipodes.