My approach:

We want to do most of the work BEFORE evaluating the two statements, so let's see what we can extract from the stem.

Since 1@352# is a multiple of 18, it must be a multiple of both 9 and 2.

Any multiple of 2 is even, so # is even. Any multiple of 9 has digits that sum to a multiple of 9, so (1 + @ + 3 + 5 + 2 + #), or (11 + @ + #) is a multiple of 9.

Since @ and # are each at most 9, we know that (11 + @ + #) can only equal 18 or 27. (It can't equal any other multiples of 9, as all of them would require @ + # to be negative or to be greater than 25).

Let's take the easier case first. If 11 + @ + # = 27, and # is even, then we must have @ = 8 and # = 8. (Any other numbers are too small.)

The other case has a few more possibilities. If 11 + @ + # = 18, and # is even, then we can find all the possibilities by trying each even value of #. We have

# = 0, @ = 7
# = 2, @ = 5
# = 4, @ = 3
# = 6, @ = 1

Combining these with the other option we found (# = 8, @ = 8), we have FIVE possibilities in total. Off we go to the statements!

S1 tells us that 5@# is a multiple of 8. Looking at our five options above, we have these possibilities:

570, 552, 534, 516, 588

Noticing that 560 (or 8 * 70) is a multiple of 8 in this range, we can quickly determine that 552 is the only multiple of 8 of the five. Hence 5@# = 552, and @ > #. SUFFICIENT!

S2 tells us that 5@# is a multiple of 6. But ALL FIVE of our numbers are multiples of 6, since all of them are even and all of them divide by 3 (check the sums of the digits). Sometimes @ > # and sometimes it isn't, so this statement is INSUFFICIENT.

Pretty neat Q!