2 cells in box 5 - R4C56 will go in R13C7. R1 has 4, 8 and 9 so R1C7 must be 5.

That was the turning point for me. I first noticed the "8" was blocked in R7 so that told me "8" is in the box on R7. So I could place "9" at R4C5. Then I noticed the "4" was blocked in box 3 R7 after that.

First, there is no border between R2C8 and R2C9 so R2C9 must be in the 40-cage. And for the second question, the remain sum in box 3 is 30 -> Box 6 need at least 2 cells to make a sum of 10. But it must be 2 cells because the 40-cage can only take 7 or 8 cells (6 cells in box 3 and 2 cells in box 6).

@@MagicOwlsVN Thanks. Right after I posted the above, I realized that the 40 cage must have a 9 in it, which places the 9 in box 6. Then r4c9 has to be in the 40-cage, and must also appear in C7 of box 3, which places the 1. After that, the rest of the puzzle is trivial. When I went back here to delete my comment - you had already responded to it! lol

click the link in the description

@@KCML82 thank you