I'm going to call the large region in this one Ken, just because.

We start with 5 freebies. The 40 must be 8x5 and can be filled.

There are five 2-cell 15s. They can be 3x5, 6+9, or 7+8. The two in column 1 must be the sums, since 3 is used; this leaves 1x2x5 for the 10 in this column. The one in column 9 must be 3x5 since 6 and 8 are used; this leaves only 1+9 for the 10 in this column, and 2+4+7 for the 13.

The 25s are interesting. They can be 1x5x5 or 9+7+9 or 8+9+8. Since they are in the same two rows, only one of the sums can be used; the other must be the product. With 5s used in both rows, the 15 in column 2 cannot be the product. The 16 above it can only be 9+7 or 2x8, so this 15 cannot be 7+8 since that leaves no way to fill the 16. So it is 9+6. This forces the sum in a 25 to be 8+9+8 to leave one 9 available for the 15. The 20 at top of row 3 can only be 4x5; this means the product is in the 25 on the right; both 25s and the 15 can be filled. The use of 5s and a 1 here also lets us fill the 10 in column 1, and then the 20.

With 5, 6, 8, and 9 used in row 3, the 14 in column 8 can only be 2x7. This and the 13 next to it can be filled, as can Ken's neck, and the 16 in column 2. The other 16 now must be 9+7, and the 6 in column 8 cannot contain a 2 or 5, so it can only be the product 6x1. In the top row, only the 6th cell can contain 8, but no other eliminations are possible now.

In column 8, the 15 now can only be 3x5, so the 21 is 4+8+9 and it can be filled in. In column 2, the upper 10 can only be 7+3, making the lower one 1+4+5. This can be filled, along with the 15 in column 9. Rows 7 and 8 both contain an 8, so the 7+8 in column 1 must go in the upper 15, but we still don't know the ordering. We have 7+8, 3+7, and 5x3 in three regions in these two rows, blocking 7 and 3 from both rows and 5 from one and 8 from the other.

The 22 must be a sum. A sum of what? Three of its cells are in a column where 5, 7, and 9 are used, and 8 and 4 are both blocked from two of these cells, so at most it can contain 8+6+3=17, so other cell must be at least 5, but 5 is blocked there also, so at least 6. Since 7 and 8 are also blocked there, this cell must be 6 or 9. That leaves only these possibilities for the 22, each in order starting from the left cell: 6+8+6+2, 6+8+2+6, 9+8+3+2, 9+4+3+6.

The 3 is rather constrained now. It could only have been 6/2/1/1, 9/3/1/1, 7-2-1-1, 8-3-1-1, 8-2-2-1, 9-4-1-1, or 9-3-2-1 from the start, but now it cannot contain an 8, nor can it contain a 9 if there are two like entries, so it can only be 6/2/1/1, 9-3-2-1, or 7-2-1-1. Each of these cases requires that the right-most cell of the 3 is either a 1 or a 9. Since the 10 in column 9 is 1+9, both these digits are blocked in that row, and the 22 can never be the cases with a 9, so the 6 and 8 can be filled in, and then both 15s and one of the 10s in rows 5-6 can be filled. The 6/2/1/1 case is not possible now, since the 22 contains two 6s in its 3 rows, and the lower 15 in column 1 also contains a 6.

In column 7, the unplaced digits in Ken's hand and foot, 3 and 4, can now be placed. Then the only place for a 3 in column 3 is the 7th cell. That forces the 3 to be 9-3-2-1, which can only go in one way, and the rest can be filled in by elimination:

325498716
584713962
198364527
269835174
876521439
732946851
643179285
951287643
417652398

Is the solution unique if a cell-group can have multiple possible operations? In this case the 6 in the upper right can be 6*1 or 6/1 (assuming I got that part right -- I broke it later on the 3).

I generally call the assignment of numbers into the cells as the "solution", without an explicit assignment of the operations required.

In this particular puzzle, I could have used a 5 or a 7 for the (1,6) clue and not had this ambiguity but I liked 6 for its broader initial possibilities I guess. Another time it was suggested that I state "each operation is used at least once" which would also resolve the operation ambiguity here but would not be a fix all for any puzzle. I'm not yet bothered enough by the operation assignment ambiguity to specifically try to avoid it.