From the question we know that X is not a multiple of 3. So when X is divided by 3, the only possible remainders are 1 and 2.
Now, Statement 1: Sum of digits is 7. If we subtract 1 from this number we get sum of digits = 6 which will make it a factor of 3. Hence we can say that when we divide this number by 3, the remainder will always be 1. Hence Sufficient.
Statement 2 is tough! we can take a sample of square numbers to verify but If i have to generalize for every number, then if x is a square which is not a multiple of 3 then its square root (say y) should also not be a multiple of 3. This gives only 2 options for y i.e 3z+1 or 3z+2. (z is some positive integer)
If we square these 3z+1 we get 9z^2 + 6z + 1 which we can clearly see will leave a remainder 1 when divided by 3.
And for 3z +2, we get 9z^2 + 12z + 4 = 9z^2 + 12z + 3 + 1 (Again remainder is 1). Hence this statement is also sufficient.