This is an interesting approach to the problem, which has been investigated to some extent, but you might have some novel insights. I'm going to translate some of this line of thought into more traditional mathematical language, and perhaps shed some light on where the Collatz classes come from.
We are looking at Collatz sequences modulo powers of 2. Let's start with 2, and then 4, then 8, etc.
Modulo 2
Every number is congruent to either 0 or 1. That means every number can be written in one of two forms:
$n_{2,0}=2j \\ n_{2,1}=2j+1$
Applying the Collatz function (let's denote it $\xrightarrow[C]{}$) to each, we see that $n_0$ drops immediately:
$2j \xrightarrow[C]{} j$, and $j$ is clearly less than $2j$. This is simply the observation that even numbers drop in one step.
The odd numbers, with form $n_1$, do not reveal anything to us, modulo 2:
$2j+1 \xrightarrow[C]{} 3(2j+1)+1 = 6j+4 \xrightarrow[C]{} 3j+2$
The result after 2 steps, $3j+2$, is not smaller than our starting value, and it could be even or odd, depending on the value of $j$, so we can't go any further. Rather, to go further, we need a larger power of 2.
Modulo 4
We've already dealt with the even numbers. Only the $2j+1$'s are a problem, and they split into two classes, modulo 4.
$n_{4,1}=4j+1 \\ n_{4,3}=4j+3$
Now we chase these through the Collatz function until we've done two divisions by 2:
$4j+1 \xrightarrow[C]{} 12j+4 \xrightarrow[C]{} 6j+2 \xrightarrow[C]{} 3j+1 \\ 4j+3 \xrightarrow[C]{} 12j+10 \xrightarrow[C]{} 6j+5 \xrightarrow[C]{} 18j+16 \xrightarrow[C]{} 9j+8$
As we can see, $n_{4,1}$ drops in 3 steps, but $n_{4,3}$ becomes uncertain after 4 steps, without dropping yet. Let's see if the next power of 2 tells us more:
Modulo 8
Our remaining class, $n_{4,3}$, splits into two classes modulo 8: $n_{8,3}$ and $n_{8,7}$. Observing these:
$8j+3 \xrightarrow[C]{} 24j+10 \xrightarrow[C]{} 12j+5 \xrightarrow[C]{} 36j+16 \xrightarrow[C]{} 18j+8 \xrightarrow[C]{} 9j+4 \\ 8j+7 \xrightarrow[C]{} 24j+22 \xrightarrow[C]{} 12j+11 \xrightarrow[C]{} 36j+34 \xrightarrow[C]{} 18j+17 \xrightarrow[C]{} 54j+52 \xrightarrow[C]{} 27j+26$
Neither of these dropped, as far as we can see modulo 8. That means we will have 4 classes, twice as many, to deal with next.
Modulo 16
Here we have $n_{16,3},n_{16,7},n_{16,11}$, and $n_{16,15}$. Sparing you the gory details, only one of the four drops here. We have $16j+3$ becoming $9j+2$ after six steps. The other three, have to push to mod 32, giving us six classes at that level.
Linear relation
What we have seen so far explains a few things in your observations. For example, numbers of the class $n_{16,3}$ drop when they have changed, in six steps, from $16j+3$ into $9j+2$. These are precisely the numbers that you identified as Class(16,9). At any rate, a linear (or affine linear) transformation turns $16j+3$ into $9j+2$, namely, $n\mapsto \frac{9}{16}n+\frac{5}{16}$. That is the line that appears in your graph.
Your non-trivial classes are the ones corresponding to larger powers of $2$, where more than one new drop appears. We haven't seen any of those yet in my analysis, so let's get a bird's eye view, and see them.
Overview
This analysis can be continued, and here are the overall results for the first few powers of 2. Note that the number of classes we start with for each power of 2 is simply the number of uncertain classes from the previous power, doubled:
$2$: 2 classes, 1 drops, 1 uncertain
$4$: 2 classes, 1 drops, 1 uncertain
$8$: 2 classes, 0 drops, 2 uncertain
$16$: 4 classes, 1 drops, 3 uncertain
$32$: 6 classes, 2 drops, 4 uncertain $\leftarrow$ This is your Class(32,27): 2 new drops $\implies$ 2 subclasses
$64$: 8 classes, 0 drops, 8 uncertain
$128$: 16 classes, 3 drops, 13 uncertain $\leftarrow$ This is your Class(128,81): 3 new drops $\implies$ 3 subclasses
$256$: 26 classes, 7 drops, 19 uncertain
$512$: 38 classes, 0 drops, 38 uncertain
$1024$: 76 classes, 12 drops, 64 uncertain
$2048$: 128 classes, 0 drops, 128 uncertain
$4096$: 256 classes, 30 drops, 226 uncertain
$8192$: 452 classes, 85 drops, 367 uncertain
$16384$: 734 classes, 0 drops, 734 uncertain
$32768$: 1468 classes, 172 drops, 1296 uncertain
Notes: The cases where there are no new drops are entirely predictable. Whenever two powers of $2$ occur between consecutive powers of $3$, the second one will have no new drops.
First, between $3$ and $9$, we have both $2^2=4$ and $2^3=8$. The latter: $8$, has no new drops. Then, between $27$ and $81$, we have both $32$ and $64$; the latter has no new drops. The only powers of $2$ that give us new drops are ones that "pass" a new power of $3$.
Additionally, the number of "uncertain" classes shown above are found in https://oeis.org/A076227
I hope that this perspective addresses your question. If I should say more, please let me know.
Best Answer
For positive integer $\ m\ $ , we need a positive integer $\ n>m\ $ with $\ 3\mid n\ $, such that the collatz-sequence beginning with $\ n\ $ contains $\ m\ $.
So, question $1$ can be answered with "yes".
Not sure about question $3$