[Puzzle 1: Let's name each cell we need to enter a number into alphabetically, left to right, top to bottom. This gives:
a+b-c=7, def=6, g+h+i=4 ad-g=1, b+e-h=4, c+f-i=2

I will only be using the following equations:
1 g+h+i=4
2 ad-g=1
3 def=6 4 b+e-h=4
5 a+b-c=7

From equation 1 we can conclude that g,h and i are 1,1 and 2 in some combination. Assuming h=i=1 would give 2+g=4, meaning g can only be 2.

We'll denote the equations where we assume g is 2 by 1g, *2g etc...

Using 2g we get ad-2=1, which gives ad=3. As three is prime a and d must be 1 and 3 in some combination.

If a=1 and d=3, then from 3g we get 3ef=6. Therefore ef=2. As 2 is prime, e and f must be 1 and 2 in some combination. We'll assume e=2 for now and if that doesn't work assume e=1.

Using 4g we now get b+2-1=4, i.e. b=3. Earlier we assumed a=1, d=3 so we can now find c from equation 5g. a+b-c=7 so 1+3-c=7, giving c=-3. This cannot be true as we are told all answers are positive whole numbers.

So we'll try e=1. 4g now gives b=4. Then 5g now gives 1+4-c=7. Then c=-2 and we have the same problem. Therefore we must have gone wrong with one of our assumptions before e=1/e=2. So if we switch to a=3, d=1 we then instead get 3g saying ef=6, meaning e is either 2 or 3.

If e=2, we have previously shown we get b=3. Plugging into 5g we get a+b-c=7, so 3+3-c=7, so c=-1. Same problem.

If e=3, we get b+3-1=4. So b=2. This gives a+2-c=7, so c=-2. Same problem.

So our problem was an assumption before even that. Our only assumption before that is g=2. You can follow this same line of reasoning for h=2 and i=2 and get the same results as with g=2. So, unless I've been incredibly stupid at some point, I think this puzzle is impossible :(](/sp)