tl;dr If $y$ represents the side length of a face, $b$ the distance between a face's center and $y$'s midpoint, $l$ the distance between the face's center and a vertex on $y$, and $\theta_0$ the angle between the face and the adjacent face across $y$, then the expressions for $r$ and $s$ to get equilateral (only for the first two chamfers) and coplanar faces are:
$ \begin{align} \theta=\frac {180-\theta_0}{2},\quad u&=b\,tan(\theta),\quad y=2\,\sqrt {l^2-b^2} \\ r&=1-\frac {y}{\sqrt {l^2+u^2} + y} \\ s&=\frac {u\,y}{\sqrt {l^2+u^2} + y} \end{align} \tag {0}$
So I managed to figure out the answer to each of the questions - what was hindering my progress was trying to figure it all out without the second image below. Once I figured that out, everything else just fell into place! The following image shows all the variables used (note that $\theta_0$ is slightly misleading this this image: the second image below is more accurate):
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/mIILll.png)
1. Relationship between $r$ and $s$ so that the vertices are coplanar
In order for the vertices to be coplanar, they all need to be on the same plane that connects the two new pentagons together, which is shown in the first image below. This was the main thing that stumped me, as I didn't know how to easily represent "coplanar vertices" without a whole lot of unknowns with vectors. This drastically simplified the workflow, and the relationship between $s$ and $r$ was quickly found (note that $\theta_0$ is the angle between the two original faces, and half of the remaining angle - $180-\theta_0$ - gets divided between them, which is why $\theta=\frac {180-\theta_0}{2}$):
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/2u7tVm.png)
$ \begin{align} \theta&=\frac {180-\theta_0}{2} \\ c&=\frac{s}{tan(\theta)} \\ r&=1-\frac{c}{b} \\ r&=1-\frac{s}{b\,tan(\theta)} \\ s&=(1-r)\,b\,tan(\theta) \end{align} \tag {1}$
2. Relationship between $r$ and $s$ so that the vertices are equilateral
This is was notably more challenging than the previous solution, and it even tricked me into thinking I had solved it a couple times - but to make things easier, I conjured up $y$ to represent the side length of the original face (not shown in the images unfortunately: this is the line perpendicular to $b$). The solution for $s$ isn't that bad, but getting $r$ was certainly not as pretty as the one before (note that only two variables from $b$, $l$, and $y$ are required, as they are connected by the expression $l=\sqrt {b^2+\frac {y^2}{4}}$)...
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/Rmzemm.png)
$ \begin{align} r&=1-\frac {d}{l} \\ d&=\sqrt {x^2-s^2} \\ r&=1-\frac {\sqrt {x^2-s^2}}{l} \\ x&=r\,y \\ r&=1-\frac {\sqrt {r^2\,y^2-s^2}}{l} \\ s&=\sqrt {r^2\,y^2 - l^2\,(1-r)^2} \\ 0&=r^2\,(y^2-l^2)+r\,(2\,l^2)+(-l^2-s^2) \\ r&=\frac {-l^2+\sqrt {l^4-(y^2-l^2)(-l^2-s^2)}}{y^2-l^2} \end{align} \tag {2.1}$
Alternatively, if we replace $r$ with $1-\frac {c}{b}=1-\frac {s}{b\,tan(\theta)}$ (instead of just leaving it in and then solving for a quadratic later), a less complex solution can be found for $r$:
$ \begin{align} r&=1-\frac {s}{b\,tan(\theta)} \\ r&=1-\frac {\sqrt {r^2\,y^2-s^2}}{l} \\ r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{b\,tan(\theta)})^2-s^2}}{l} \end{align} \tag {2.2}$
3. Relationship between $r$ and $s$ so that the vertices are coplanar and equilateral
Given that I need both coplanar and equilateral faces here, I simply used a system of equations using both the previous solutions and solved for $r$ and $s$ independently. There was a lot of simplifying done and the process was not pretty in the least, so for this answer I just laid out the groundwork and then skipped to the solution - let me know if you wish to see the entire process, or even if you find an easier/better way to do this! For the sake of clarity, I replaced all instances of $b\,tan(\theta)$ with $u$:
Solving for r (using the upper root):
$ \begin{align} s&=(1-r)\,u \\ s&=\sqrt {r^2\,y^2-l^2\,(1-r)^2} \\ (1-r)\,u&=\sqrt {r^2\,y^2-l^2\,(1-r)^2} \\ (1-r)^2\,u^2&=r^2\,y^2-l^2\,(1-r)^2 \\ &... \\ r&=\frac {-(l^2+u^2)+\sqrt {(l^2+u^2)^2-(y^2-l^2-u^2)\,(-l^2-u^2)}}{y^2-l^2-u^2} \end{align} \tag {3.1}$
Solving for s (using the lower root):
$ \begin{align} r&=1-\frac{s}{u} \\ r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{u})^2-s^2}}{l} \\ 1-\frac{s}{u}&=1-\frac {\sqrt {y^2\,(1-\frac {s}{u})^2-s^2}}{l} \\ \frac{s^2}{u^2}&=\frac {y^2\,(1-\frac {s}{u})^2-s^2}{l^2} \\ &... \\ s&=\frac {u\,y^2-\sqrt {u^2\,y^4-(y^2-l^2-u^2)\,(u^2\,y^2)}}{y^2-l^2-u^2} \end{align} \tag {3.2}$
Thanks to WolframAlpha, a much simpler solution could be produced if all the values are assumed to be positive (given that we will never have negative distance or angles, this is a valid assumption):
$ \begin{align} s&=\frac {u\,y}{\sqrt {l^2+u^2} + y},\;u=b\,tan(\theta),\;y=2\,\sqrt {l^2-b^2} \end{align} \tag {3.3}$
And given that $r=1-\frac {s}{u}$, this gives us a nice solution for $r$ as well:
$ \begin{align} r&=1-\frac {y}{\sqrt {l^2+u^2} + y},\;u=b\,tan(\theta),\;y=2\,\sqrt {l^2-b^2} \end{align} \tag {3.4}$
In applying this to my dodecahedron (the magnitude of each vertex being $1$), I can now show that the solutions I discovered are correct (credit for the values for $r$ and $s$ goes to Blue's answer - note that these values are all approximations, rounded to the fourth digit-place):
Given: $b=0.4911,\quad l=0.6071,\quad y=0.7136,\quad \theta=\frac {180-116.5650}{2}=31.7175,\quad u=b\,tan(\theta)=0.3035$
Unknown: $r=0.4875,\quad s=0.1556$
1. Relationship between $r$ and $s$ so that the vertices are coplanar
$ \begin{align} r&=1-\frac{s}{b\,tan(\theta)}&=1-\frac{0.1556}{0.4911\,tan(31.7175)}&=.4873 \\ s&=(1-r)\,b\,tan(\theta)&=(1-0.4875)\,0.4911\,tan(31.7175)&=.1556 \end{align}$
2. Relationship between $r$ and $s$ so that the vertices are equilateral
$ \begin{align} r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{b\,tan(\theta)})^2-s^2}}{l} &=1-\frac {\sqrt {0.7136^2\,(1-\frac {0.1556}{0.4911\,tan(31.7175)})^2-0.1556^2}}{0.6071}&=.4877 \\ s&=\sqrt {r^2\,y^2 - l^2\,(1-r)^2}&=\sqrt {0.4875^2\,0.7136^2 - 0.6071^2\,(1-0.4875)^2}&=.1556 \end{align}$
3. Relationship between $r$ and $s$ so that the vertices are coplanar and equilateral
$ \begin{align} r&=1-\frac {y}{\sqrt {l^2+u^2} + y}&=1-\frac {0.7136}{\sqrt {0.6071^2+0.3035^2} + 0.7136}&=.4875 \\ s&=\frac {u\,y}{\sqrt {l^2+u^2} + y}&=\frac {(0.3035)\,(0.7136)}{\sqrt {0.6071^2+0.3035^2} + 0.7136}&=.1555 \end{align}$
Update
These equations may work for future subdivisions/chamfering as well, however faces will not continue to be equilateral (while still being coplanar). This is to be expected, however an issue arises of how to uniformly scale and translate the vertices of a face if it is not equilateral (given that $b$, $l$, and $y$ may vary per vertex). As such, my chamfering process would need to be updated so that it works with such faces.