This is part of a bigger problem, but I just wanted to ensure the work I did in this portion is correct.
I'm using the generators $r$ and $f$, where $r^n=e$, $f^2=e$, and $rf=fr^{n-1}$.
Here's my proof:
Consider the group $D_{2n}$ where $n \equiv 0 \text { mod }2$ and the element $r^{n/2}$. Then
$r^{n/2}f=r^{n/2-1}(rf)=r^{n/2-1}fr^{n-1}=r^{n/2-2}(rf)r^{n-1}=r^{n/2-2}fr^{2(n-1)} = \dots = fr^{n/2(n-1)}=f\left(r^{n-1}\right)^{n/2}=f\left(r^{-1}\right)^{n/2}=f\left(r^{n/2}\right)^{-1}=fr^{n/2}.$ As needed.
Also, as a notational note, $D_{2n}$ is the dihedral group of order $2n$. (So $D_6$ is the dihedral group of a triangle).
Your proofs seems fine. You can also note that, because $r^n= 1$ we have that $r^{n/2} = r^{-n/2}$ and since $rf = fr^{-1}$ we have that $r^{n/2}f = fr^{-n/2} = fr^{n/2}$.