Are the following statements equivalent to the parallel postulate?

Both (i) and (ii) are strictly weaker than the parallel postulate. Probably the easiest way to see this is using coordinates. Given any Pythagorean ordered field $F$ and a convex subring $F_0\subseteq F$, the full subplane $F_0^2$ of the coordinate plane $F^2$ is a Hilbert plane which will always satisfy (i) and (ii), and will not satisfy the parallel postulate unless $F_0=F$. (If $F$ is non-Archimedean, you can always find such an $F_0$ which is not all of $F$, for instance by taking $F_0$ to be the set of all "finite" elements of $F$, i.e. elements bounded by integers.) Conversely, every Hilbert plane satisfying the parallel postulate is isomorphic to the coordinate plane over some Pythagorean ordered field, and so will satisfy (i) and (ii).

Here is a proof that $F_0^2$ satisfies (i) and (ii). For (i), suppose we have a non-vertical line with equation $y=ax+b$. If the slope $a$ is in $F_0$, then $ax\in F_0$ for all $x\in F_0$, and thus we must have $b\in F_0$ for the line to have any points in $F_0^2$ at all, in which case the line intersects every vertical line. On the other hand, if the slope $a$ is not in $F_0$ (or if our line is vertical), then $a$ must be infinitely large (since $F_0$ is a convex subring of $F$, it contains all elements which are bounded by an integer), so $a^{-1}$ is infinitesimal and is in $F$. Rewriting the equation as $x=a^{-1}(y-b)$, the argument from the first case now shows that our line intersects every horizontal line.

Thus, every line in $F_0^2$ either intersects every vertical line or intersects every horizontal line. Given three lines, we may thus assume without loss of generality that two of them intersect every vertical line. Now take any point on the third line, and the vertical line on that point will be a transversal to all three lines. (Unless the third line is itself vertical, in which case you can take its intersection with the first line and draw the line from it to any other point on the second line, choosing the latter point so that the line is not vertical.)

For (ii), suppose we have four (or indeed any finite collection of) points in $F_0^2$. Choose $a<b$ in $F_0$ such that both coordinates of all four points are strictly between $a$ and $b$. Then the right triangle with vertices $(a,a)$, $(a,2b-a)$, and $(2b-a,a)$ contains all four points. (Note that this argument actually requires only that $F_0$ is a subgroup of $F$, not a subring as was needed for (i).)