yeah, so you should definitely rephrase the problem by getting rid of the fractions first. you can kill them all by multiplying through by 12, the common denominator of all the fractions:
2k + 3m = t
this is the base equation from which we'll be working to answer the problems.

you can also rephrase the question. if t and 12 have a common factor greater than 12, that's just another way of saying that something that goes into 12 (besides 1) also goes into t. so, here's a rephrase of the prompt question:
does 2, 3, 4, 6, or 12 go into t?
since 4, 6, and 12 are just compounds of the prime factors 2 and 3, it's actually redundant to ask whether those numbers are factors of t. therefore, we can rephrase the question down to its most distilled essence:
does 2 or 3 go into t?

-- statement (1) --
if k is a multiple of 3, then we can write k = 3n, where n is an integer. therefore, we have
2(3n) + 3m = t
6n + 3m = t
since 6n and 3m are both multiples of 3, it follows that 3 goes into t.
sufficient.

-- statement (2) --
if m is a multiple of 3, then we can write m = 3n, where n is an integer. therefore, we have
2k + 3(3n) = t
2k + 9n = t
these (2k and 9n) don't have a common factor, so this is insufficient.
if you want proof of insufficiency, go ahead and plug some easy numbers: if k = n = 1 (which would actually mean k = 1 and m = 3), then t = 11, which doesn't share a common factor with 12. if k = 3 and n = 1 (which would actually mean k = m = 3), then t = 15, which shares the common factor 3 with 12.
insufficient.

answer = a