For solving: https://sudokupad.app/zk5mc10cfr

1. Box 5 looks like a nice place to start! All but two cells are thermometer tips with a minimum value of 3. Since R5C5 is even, it's 2, while R6C4 is 1.

2. In box 8, R9C5 is now 1. In box 6, 1 must be in either R5C8 or R5C9. Because of that, in box 4, 1 must be in either R4C1 or R4C2.

3. Let's pencil in thermometer and even candidates. Some thermometer cells have as many as six candidates, but none should have seven.  You don't have to pencil everything in, but it might make it easier to spot some things. For example, where can 9 go in column 3, or column 7, or row 3, or row 7? In each case, the answers involve two cells which are on either row 2 or 8 or column 2 or 8. The result is a pair of overlapping "X-Wings" that interact such that placing a 9 in any of the eight cells will immediately resolve 9s in boxes 1, 3, 7, and 9. Since these eight cells account for all the 9s in those four boxes, as well as all the 9s in rows 2, 3, 7, and 8 and columns 2, 3, 7, and 8, the tip of thermometer in R8C5 is no longer 9, and we can place 9 in R9C4. That eliminates 9 from two more thermometer tips and whittles down some thermometer candidates on affected arrows.

4. Where can 3 go in box 5? To answer that, consider where 3 can go in column 4. Between given and placed digits and even cells, there are only three options, and one of those would require a second 2 in row 5, so there are only two options. Whether 3 is R3C4 or R5C4, it cannot be on R4C5. So there are two options for 3 in box 5: R5C4 or R6C5. Since R6C3 shares a row or thermometer with both of those, it cannot be 3.

5. Same question, rotated, for 5 in box 5. Again, between given digits and even cells, there are only four options for 5 in row 4, and one of those would require duplicate digits in box 2. Whether 5 is R4C4, R4C5, or R4C7, it cannot be on R5C6. So there are three options for 5 in box 5: R4C4, R4C5, and R5C4. Since R3C4 shares a column or thermometer with all three of those, it cannot be 5.

6. Where does 8 go in box 4? The 9 in row 9 had an effect on the thermometers in box 4, so the given 8 forces 8 into row 4, where as a "pointing pair" of cell, it points across and eliminates 8 as a candidate in five other cells on that row. This further whittles down thermometer cells in box 3 and box 4.

7. Now let's revisit where 3 can go in box 5. Can it still be in R5C4? Consider it as part of a group with R5C2, R5C3, as well as R6C1 and R6C3. The four cells in box 4 are made of five candidates: 2, 3, 4, 6, and 7. Normally five candidates for four cells wouldn't be helpful. But if R5C4 is 3, it eliminates both 3 (by row) and 7 (by thermometer) from that pool, leaving only three digits for four cells in box 4. So R5C4 cannot be 3, and R6C5 is 3. Since we have a given 3 in column 6 and just placed a 3 in column 5, we can also place a 3 in R3C4.

8. The 3 in R6C5 places 2 and 1 down that thermometer, and now R3C6 is even, but sees 2, 4, and 6, so it is 8. Similarly, R7C4 is even but now sees 2, 4, and 8, so it is 6. It is difficult to choose which of many follow-on steps to pursue next, but let's stick with even cells, at least: R1C4 now sees all even digits but 2, and both R7C9 and R9C6 see every even digit but 4. Finally, R4C9 can't be 8 because of the "pointing pair" of cells on that row in box 4, so it is 2.

9. That 3 in R6C5 also has a strong effect on the thermometer that starts at R2C5. Since the bulb can no longer be 3 (nor 1, 2, or 4), the lowest it can be is 5, so R3C5 is 7 (because of the given 6 and placed 8 in that row), and R4C6 is 9 because of the same "pointing pair" of 8s in that row in box 4.

10. That places 9 in R1C5 and R5C1, which in turn places 9 in R6C9. It also eliminates 9 from the tips of three thermometers in box 5, and two of them also lose 8, while the other also loses 7. In fact, there's only one cell in box 5 that can still be 8, so R5C4 is 8.

11. Also in box 5, the thermometer cells in column 6 can't be 4, so they are a 6-7 pair. That removes 7 from R4C4, making it 5 and R4C5 a 4.

12. We can finish out column 5 starting with R7C5 which can only be 5, making R8C5 8 and R2C5 6. There's only one unresolved cell in column 4, so R8C4 is 7.

13. Revisiting even cells, R4C3 now sees 2, 4, and 6, and so is 8. R8C8 now sees 4, 6, and 8, and so is 2. Because of that, R2C8 now sees 2, 4, and 6, and so is 8, after which R1C7 is 4, after which R6C7 is 8. R2C2 can only be 2, which means R3C1 can only be 4.

14. We've eliminated all but one candidate for many cells: R3C3 is 1, R3C7 is 2, and both R4C7 and R7C3 are 3. After placing the 3 in box 6, now R5C7 can only be 5.

15. Consider row 5: R5C2 is the only cell into which 3 can go. Now consider row 6: R6C1 is the only cell into which 2 can go. In column 3, the only cell into which 2 can go is R9C3. That places all the 2s in the puzzle.

16. In box 2, we need only 1 and 5. The given 5 in row 1 forces the 5 into R2C6, putting a 1 in R1C6.

17. Let's resolve the "overlapping X-Wings" from step 3. We can use row 3, row 7, or column 7, each of which is missing only two digits, with one of the cells sharing a column or box with a given digit. So the 9s go in the more-counter-clockwise (or anti-clockwise) side, in R7C8, R2C7, R3C2, and R8C3, and all 9s are placed. It is now possible to circle around again and fill in the other half of each pair: R8C7 is 6, R3C8 is 5, and both R2C3 and R7C2 are 7.

18. Now the final even cell in R8C2 must be 4.

19. The missing digits in row 4 are 1 and 7, which our recently-placed 7 in column 2 resolves. The other two cells in box 4 can be resolved by realizing 6 is not between 3 and 5. The 6 in R6C3 also resolves the outstanding pair in box 5, while the 4 in R5C3 resolves the triple in box 6.

20. What remains are the corners. Column 2 needs 6 and 8, which a given 8 in box 7 resolves. Column 9 needs 3 and 7, which a given 7 in box 9 resolves. Row 8 needs 1 and 5, which a placed 1 in box 9 resolves. Row 2 needs 1 and 3, which placed 1s resolves. Finally, the four corners are left as an exercise for the reader.