Grids with trominoes

Question

Let's have two 8x8 grids. By visual inspection we see they are filled with trominoes of three different colors. There are 7 trominoes of each color. On the grids the trominoes are not allowed to touch anywhere side to side with the same color. As you can see the position A4 has two pairs marked with red ink and the position B2 has 8 pairs. Can you rearrange the trominoes on the grids, so that you have one pair on the grid with position A4 empty and 9 pairs on the grid with position B2 empty? A pair is a 2X3 or 3X2 rectangle made with trominoes.
     
Transcription of left-side image (B = blue, R = red, T = teal)
|B |R |R |T |T |B |T |T |
|B |B |R |T |B |B |R |T |
|R |R |B |B |T |R |R |B |
|A4|R |B |R |T |T |B |B |
|B |B |T |R |R |B |R |R |
|B |R |T |T |B |B |T |R |
|T |R |R |B |R |T |T |B |
|T |T |B |B |R |R |B |B |

Transcription of right-side image:
|T |T |R |R |B |B |R |R |
|T |B2|R |T |T |B |T |R |
|B |B |T |B |T |R |T |T |
|B |T |T |B |B |R |R |B |
|T |B |B |R |R |T |B |B |
|T |T |B |R |T |T |R |R |
|R |R |T |B |R |R |B |R |
|R |T |T |B |B |R |B |B |

theonetruepath · Accepted Answer

Answers to parts 1 and 2.
Part 1:

Part 2:

My method.
I used my solver with a few new tweaks as follows:

I already had "use a maximum of N no-touch colours" which I set to 3. I also set what I call "non-strict neighbours" to allow the same colours to touch at corners.
This gave about 1.2 million solutions for the first one which I call 'problem A'. So I added some "Max per colour" code and set it for 7 colours. Now we are at around 120k solutions. Some of these are the same layout with a different colouring.
I turned on "similar pattern rejection" which tracks solutions found and rejects any which have the same layout, independent of colouring. Now we are at 11243 tilings found. This means there are on average around 10 ways to colour each tiling. There are other tilings not 3-colourable as well but I already filtered those out.
I added code to count the number of occurrences of a new piece being added sharing exactly N neighbouring cells with a piece already placed. I set N to 3. I believe all 2x3 rectangles formed from a pair have this characteristic, and all non-rectangular placements do not. I also specified a maximum and minimum count for these occurrences. Now I get tiling counts as follows for number of rectangles 0 through 10:
0-0, 1-2, 2-18, 3-36, 4-85, 5-282, 6-1362, 7-1046, 8-895, 9-1683, 10-5834.
And if you sum these you get 11243 which is encouraging.

However... When I do the same for 'problem B' the situation is complicated by the fact that the empty square is in a symmetric position, which means I get double the tilings unless I turn on a different sort of symmetry rejection, which is not proven with the new code I added. So I just divide the counts by 2 instead, then I get
0-1.5,
1-2,
2-26,
3-5,
4-54,
5-127,
6-598,
7-126,
8-191,
9-0,
10-3007.
Which sums to 4137.5. Solving without regard to rectangles gives 8275/2=4137.5 as well. Alert readers will question the half tiling which is caused by a symmetric tiling with zero rectangles, apparently the only such tiling. The correct counts are 2 tilings with zero rectangles and 4138 total.
Now I turned my solver onto the other posted problem like this which I call "Problem C", although I believe it was posted earlier. The empty square is asymmetrically placed which avoids symmetry rejection issues, and I get counts as follows:
0-2,
1-4,
2-44,
3-10,
4-121,
5-225,
6-1185,
7-252,
8-373,
9-0,
10-5911.
Sum is 8127. When I run the solver to find tilings without regard to rectangles, I get 8127 as well.
If there is indeed a 9-rectangle tiling for "Problem B" then I have a bug affecting both ways of counting, which is not that unlikely with the new code I added yesterday.
I fixed the symmetry rejection in the symmetric cases and ran the solver on all possible empty cell positions. Using OP's naming for empty square position I get these counts, all match with the counts from the solver when omitting the 'find a specific number of rectangles' code. When omitting the 3-colouring requirement for case B2 I still get zero 9-rectangle tilings.

Rectangles
A4
B2
A2
A1
A3
B3
B4
C3
C4
D4

0
0
2
2
1
0
1
0
0
4
1

1
2
2
4
2
0
0
0
0
0
0

2
18
26
44
26
14
24
22
4
21
16

3
36
5
10
47
46
19
16
21
50
19

4
85
54
121
83
247
65
129
110
266
51

5
282
127
225
411
296
201
219
247
401
249

6
1362
598
1185
1181
1148
532
1146
297
837
504

7
1046
126
252
912
2014
406
551
693
1514
421

8
895
191
373
1375
1631
178
164
2268
1055
495

9
1683
0
0
1249
3498
1370
1264
0
3412
941

10
5834
3007
5911
2936
0
0
3997
0
0
1900

sum
11243
4138
8127
8223
8894
2796
7508
3640
7560
4597

count
11243
4138
8127
8223
8894
2796
7508
3640
7560
4597

Grids with trominoes

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Rectangles	A4	B2	A2	A1	A3	B3	B4	C3	C4	D4
0	0	2	2	1	0	1	0	0	4	1
1	2	2	4	2	0	0	0	0	0	0
2	18	26	44	26	14	24	22	4	21	16
3	36	5	10	47	46	19	16	21	50	19
4	85	54	121	83	247	65	129	110	266	51
5	282	127	225	411	296	201	219	247	401	249
6	1362	598	1185	1181	1148	532	1146	297	837	504
7	1046	126	252	912	2014	406	551	693	1514	421
8	895	191	373	1375	1631	178	164	2268	1055	495
9	1683	0	0	1249	3498	1370	1264	0	3412	941
10	5834	3007	5911	2936	0	0	3997	0	0	1900
sum	11243	4138	8127	8223	8894	2796	7508	3640	7560	4597
count	11243	4138	8127	8223	8894	2796	7508	3640	7560	4597