Looking at the first grid
The product of the middle column is 21, which is only divisible by 1, 3 and 7. Therefore the numbers in the middle column had to be 1, 3 and 7.
The product of the left column and the middle row are multiples of 10 therefore they had to have 5 as one of their numbers. The only square they share is the middle square of the left column so that is where I put the 5. I concluded that the only number multiplied by 5 that would help you to get to 40 was 1. So I put 1 in the middle square.
Because 24 is not divisible by 7 and the bottom row has a high product I put 7 in the middle of the bottom row, leaving the 3 in the middle of the top row.
I then turned my attention to the middle row and calculated that 1x5=5 and 5x8=40, so I put the eight in the right-hand square of the middle row. I then concluded that 9 was in the right-hand square of the bottom row because the products of the right-hand column and bottom row were high.
From there I calculated all the missing numbers.
2x3x4=24
5x1x8=40
6x7x9=378
2x5x6=60
3x1x7=21
4x8x9=288
Looking at the next grids
For the first grid I calculated that the 5 would go n the right-hand square of the middle row as both the right-hand column and middle row both have a product 10 times tables. However I could not work out a solution to the red grid, as there are not 3 ways in making 24.
On the second grid I worked out that the 5 needed to go in the left-hand column on the bottom row because 90 and 40 are both multiples of 10. From then onwards I calculated the rest.
For the last grid I calculated that I’d have to put a 5 in the left-hand column on the middle row and from then onwards I calculated the rest.