Your answers are right for most of them.
For (d), the notion that "$a = \pm b$" is somewhat ambiguous and can be confusing, but usually means that "$a = b$ or $a = -b$". You might more compactly notate this as $|a|=|b|$ as a result. In which case, your desired result pretty much follows from the fact that $=$ is itself an equivalence relation:
- Reflexivity: Holds, as $|a|=|a|$ trivially
- Symmetry: Holds, as $|a|=|b|$ implies $|b|=|a|$
- Transitivity: Holds, as $|a|=|b|$ and $|b|=|c|$ implies $|a|=|c|$
(f) is a confusing one, because it depends on how one defines lines, parallelism, and intersections. It might seem a little pedantic, but it is noteworthy.
Suppose you define two lines to be parallel if they have no intersection point. Then a line $\ell$ is of course not parallel to itself: it intersects itself infinitely many times. Thus $\ell \not \sim \ell$ by this relation, and thus no reflexivity.
Suppose you define two lines to be parallel if and only if the shortest distance between them is constant, no matter where you measure from. Then $\ell \sim \ell$ after all, because that constant distance is $0$. You can go on to show symmetry and transitivity as well.
Since we're in a plane, we could define lines by their slope and $y$-intercept on a graph. Let $\ell = ax+b$ and $m=cx+d$. Thus, $\ell \sim m$ if $a=c$ and $b \ne d$, presumably. In which case, $\ell \not \sim \ell$. But then again, this choice of $b \ne d$ is somewhat arbitrary insofar as it just means we can only look at distinct lines; if you remove that condition, then of course $\ell \sim \ell$.
It's all about definitions in the end; you have to be careful with how you phrase things. (Or, rather, your source should have more carefully defined the notion of "parallel," but I digress.)