To solve this puzzle, if one uses the traditional approach, chances are one has to apply quite a few advanced fishy techniques to solve it. I am going to introduce a completely different solving approach which I name as "label solving" to tackle this puzzle.

1. The first step is to ignore the actual given clues, but to fill in one area of the grid where we know must contain 9 different digits. In this case I choose the 9 uncoloured cells which are not covered by the 8 given "45-cages", and label these cells with 9 different letters:

2. Directly from this layout, we can find 4 hidden singles and 4 hidden pairs for the 4 green regions:

3. Also, because R3C3 sees R2C12345, it has to map to one of R2C6789, and since R1C12=R2C78, R3C3 must map to one of R2C69. Similar to the rest of R37C37:

4. This is the most tricky part: from the 8 initial givens (which we have been ignoring till now), since they are all having different values, so as the labels covering those exact same cells. We can actually consider these 8 cells as a hidden disjoint "zero cage".

Obviously the label E cannot appear on those 8 cells, so the rest {A,B,C,D,F,G,H,K} must each appear once on them. 4 of them {A,B,H,K} must appear in D/2378, leaving the other 4 {C,D,F,G} to be in D\28+D/46:

5. Applying CPE on this hidden cage, we have R2C4 seeing all F's and R8C6 seeing all D's:

6. As a result we have 2 hidden pairs on R2 & R8 (R2C25={FG}, R8C58={CD}):

7. R1C3, seeing R2C134 as well as R1C129 & R23C2, have only 1 possibility left. Ditto for R9C7:

8. Next we have 2 hidden triples on N2 & N8 (R3C456={ABD}, R7C456={FHK}):

9. Two simple row-cage intersections on two of the bloom petals:

10. Another two row-cage intersections on the remaining two bloom petals:

11. A series of hidden singles emerge after the previous step:

12. With R3C9 seeing R134C8 there remains only 1 possibility left. Ditto for R7C1, resulting in a series of singles:

13. More CPEs. R4C6 sees all F's of a petal while R6C4 sees all D's of another:

14. Finally, two more simple row-cage intersections on two petals clean it up:

15. This is the "labelled solution":

To obtain the answer for the original puzzle, just convert all labels back to numbers corresponding to the initial givens. (F=1, B=2, K=3, G=4, C=5, A=6, H=7, D=8, E=9)

The thing I like about this approach is how we treat the initial givens as a "hidden cage" and a "hidden cage" as initial givens. Amazingly this simple twisting of concepts makes the solving so much easier. I am sure other difficult puzzles with 45-cages might be tackled more easily with this approach. Will be thankful if someone would give more examples.