1 - sufficient. 3 @ 2 > 3 implies @ is either + or *.
For both + and *, (6@2)@4=6@(2@4)

2 - insufficient. @ can be either * or /.
If we use /, (6/2)/4 != 6/(2/4)
If we use *, (6*2)*4 = 6*(2*4)

Hence A

So the first has two arithmetic signs + and x, but if they both give the same answer on both sides of the equation then the answer is sufficient? so for future, how do I approach these kinds of questions?

Even if there are 2 possible signs, as long as they give me the same number on both sides of the equation then its suffient, right?

If the operation [@] is one of the four arithmetic operations addition, subtraction, multiplication and division is (6@2)@4=6@(2@4)

I. 3@2>3

II. 3@1=3

3@2>3. Thus @ for sure can't be division or subtraction. But it can be either addition or multiplication in in both cases the result would be greater than 3.
Now if @ is addition then (6@2)@4 would be (6+2)+4 = 6+2+4=12; and 6@(2@4)=6+(2+4)=6+2+4=12
If @ is multiplication then (6@2)@4=(6*2)*4=48; and 6@(2@4)=6*(2*4)=48
Hence in both cases (6@2)@4=6@(2@4)
Hence alone sufficient

3@1=3. Here @ can be multiplication or division. If @ we know that the two sides of the equation would be the same. But in case of division
(6/2)/4 is not equal to 6/(2/4)
First value = 3/4
Second value = 12

Hence Answer is A

I get it now, thanks