On Target
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For solving: https://sudokupad.app/mvyhrk2eos

1. The hollow red parity lines that cross in the center of the grid intersect with the solid green whisper lines at four points. The green whisper lines mean that the two points on one red parity line are low digits, while the other two points on the other red parity line are high digits. With that in mind, are those points odd or even?
1a. There are several ways to determine the points cannot be even: all four points are different digits, so if they are even, there is no possible value for R5C5. Or: one pair must contain a 4, and the other pair a 6, both in between two cells on a green whisper line, which would require the 4 to be surrounded by 9s, and the 6 to be surrounded by 1s, all of which is impossible.
1b. So the four intersection cells are odd, and the alternating cells on the red lines are even.
1c. That will give you four even cells in row 5 and column 5, so the "extra" cells beyond the red lines are also odd. Of course, R1C5 and R5C1 cannot be 1, R1C5 and R5C7 cannot be 7, and 5 doesn't belong on green whisper lines, and so isn't in R3C5, R5C3, R5C7, R5C9, or R7C5.
1d. Since all four intersection points must be different digits, and they are now a 1379 virtual quadruple, R5C5 is 5, and the four corner cells of box 5 are 1379, settling the parity of two four-cell red parity lines.

2. The just-placed 5, combined with green whisper cells in box 6, forces R4C9 to be 5. Both placed 5s, combined with green whisper cells in box 4, force R6C1 to be 5.
2a. Continuing with 5s: the green whisper cells in boxes 2 and 8, combined with the 5 in the center of the grid (or the red parity line, if you prefer), means that 5 in box 2 is in row 1, and 5 in box 8 is in row 9.
2b. Now those pointing pairs of 5s rule out the top and bottom rows in the corner boxes, while the placed 5s in columns 1 and 9 rule out the left and right columns in the corner boxes, leaving only four possible cells in each for a 5 to go. In boxes 3 and 7, one of the four is even, and two are on a green whisper line, so R2C8 is 5 and R8C2 is 5.
2c. Finally, those placed 5s have stripped away three of the four possible cells in boxes 1 and 9, so R3C3 is 5 and R7C7 is 5.

3. Now we need to figure out which parity line crosses the whisper line at the low digits, and which crosses at the high digits.
3a. We can start by observing that if the row 5 line crosses at the high digits, then R5C7 could only be 9, so R5C3 cannot be 9.
3b. If the column 5 line crosses at the high digits, then the intersection points would be a 79 pair, which would force R1C5 to be 3 and R9C5 to be 1. In that case, the 1 in R9C5 would keep 7 from being an option in box 8, since a 7 on a green whisper line requires two digits below three. So we can eliminate 7 from R7C5 and 9 from R3C5, as well as 3 from R9C5.
3c. If the column 5 line instead crosses at the low digits, then R1C5 would be 9 and R9C5 would be 7. In that case, the 9 in R1C5 would keep 3 from being an option in box 2, since a 3 on a green whisper line requires two digits above seven. So we can eliminate 3 from R3C5 and 1 from R7C5, as well as 9 from R9C5.
3d. Now R3C5 and R9C5 are a 17 pair, and R1C5 and R7C5 are a 39 pair.
3e. The same logic applied to row 5 puts 37 in R5C1, 17 in R5C3, 39 in R5C7, and 19 in R5C9. We get a little more information in row 5 than we got in column 5: the 7 in box 4 is in either R5C1 or R5C3, and the 9 in box 6 is in either R5C7 or R6C9.
3f. That's important, because if R5C7 were 3, that would require it to have 8 and 9 above and below it, but 9 would be in R5C9, and can't also be in column 7. So R5C7 is 9, which settles the green whispers line in terms of highs and lows.
3g. To be clear: R5C1 is 3, R5C3 is 7, R5C7 is 9, and R5C9 is 1. R1C5 is 9, R3C5 is 1, R7C5 is 3, and R9C5 is 7.

4. The 7 and 3 on the solid green whisper lines have very limited neighbors: 7 is surrounded by 12, and 3 is surrounded by 89. That means R7C3 sees both a 2 in the column and an 8 (and 6) in the row, so R7C3 is 4.

5. Now there's a 12345 quintet and a7 in column 3, so the remaining three cells in the columns are 689, although R1C3 cannot be 9 and R8C3 can't be 6.
5a. That settles the high/low of the green whispers line in box 7, so R7C2 is 12, R6C2 is 689, and R8C4 is 124.

6. The green whispers line in box 1 is also settled, because R2C3 cannot be 6, since that would require both adjacent cells on the line to be 1, and neither can be. So R2C3 and R4C2 are 89, while R2C4 and R3C2 are 234.

7. There's now an 89 pair in column 3, so R1C3 is 6. Then R2C2 is odd, so 379, and R3C1 is even, so 248.

8. Box 7 is interesting. Since R9C1 can only be 2 or 8, the whisper cells in the box must contain either 1 or 9 or both.
8a. More to the point, though: the only place for 7 in box 7 is R8C1.
8b. With that placed, R2C2 is 7.

9. Since we're looking at four-cell green whispers, let's consider where 1 goes in box 3. The only option is R2C7.
9a. That settles the line: R2C6 is 68, R3C8 is 89, and R4C8 is 23.

10. Now there's a 234 triple in box 6, in R4C7, R4C8, and R6C7. That means there's a 678 triple in R5C8, R6C8, and R6C9.

11. Since R6C8 is high, R7C8 is low, 12, R8C7 is 68, and R8C6 is 12.

12. Now there's a 12 pair in row 78, so R7C9 is 7, and so is R6C8.

13. In fact, there are a number of simple deductions available now: Since R8C8 must be even, R8C9 is 3.
13a. Since 4 on a green whispers line must be next to a 9, and 9 is not available in R3C6, R6C7 is 4.

14. Now R6C5 and R6C9 are a 68 pair, so R6C2 is 9.

15. On row 6, R6C4 and R6C6 are a 13 pair, so R6C3 is 2, R4C3 is 1, R4C2 is 8, R4C1 is 4, and R5C2 is 6.

16. Now in box 7, there's a 12 pair in column 2, so R9C1 is 8 and R8C3 is 9.

17. In box 1, R2C3 is 8, R3C1 is 2, and R2C1 is 9. Since 8 on a green whisper is not far enough away from 4, R3C2 is 3 and R1C2 is 4.

18. More singles: R5C8 is 8, R6C9 is 6, R6C5 is 8, and R4C5 is 6.

19. After that, R8C8 is 6, then R8C7 is 8, then R3C7 is 6, R3C8 is 9, and R2C6 is 6.
19a. Now R9C6 is 5, R9C4 is 6, R1C4 is 5, and R1C6 is 23.
19b. At this point, R1C9 is 8, R7C8 is 1, R7C2 is 2, R9C2 is 1, R9C7 is 2, and R9C9 is 9.

20. Continuing, R4C7 is 3, R4C8 is 2, R1C8 is 3, R1C6 is 2, R2C5 is 4, R2C4 is 3, R2C9 is 2.
20a. Now R3C9 is 4, R5C6 is 4, R5C4 is 2, R6C4 is 1, R6C6 is 3, R8C6 is 1, R8C5 is 2, and R8C4 is 4.

21. Now we have a 789 rotating triple in columns 4 and 6. To resolve it, consider either that a 3 on a green whispers line needs at least an 8 as a neighbor, or that a 4 on such a line needs a 9 as a neighbor.