First I looked at 5,7,9, because that appears in three hints (1), (2) and (4).
Obviously it is not possible for none to be in the code (each hint has three correct digits). If all three are correct, we can exclude 1,2,3,4,8, but then (5) is impossible. If only one is correct, then the other two digits in those hints must be correct, so we end up with 1,2,3,4,8 and one out of 5,7,9 being correct. That is six digits for a five digit code and thus impossible. As a consequence exactly two out of 5,7,9 are correct.
Next, an observation: (1) and (4) as well as (3) and (6) differ in only one digit (and order). Since all hints have the same amount of correct digits the differening digits are either both correct or both incorrect. This applies to the pairs 1,4 and 6,8.
Next up we are left with three digits we did not consider yet: 0,2,3. Those three all appear in (3) and (6). We know that two out of 5,7,9 are correct and that 1,4 and 6,8 only appear if the other one appears, so in any case we will have an even number of correct digits, so the number of correct digits in 0,2,3 has to be odd, since 5 is odd.
If all three are correct we know five correct digits, so all possible digits in the code: 0,2,3, two out of 5,7,9. Requireing that in each hint only three digits are correct, we can eliminate 9 due to (3). 0,2,3,5,7 proves to ba a valid set of digits. Now for positions: Since in each hint all digits are wrongly placed we have to look at where a digit does not appear. This leaves only one possibility for 7 and 3: xxx37. Now we can fill in xx037 and now neither 2 or 5 can appear in the second position, making the assumption wrong. Thus only one out of 0,2,3 is correct.
In (6) and (3) only one out of 0,2,3 is correct, so the other two digits have to be correct. This leaves us with 6,8,9. Consequently both 1 and 4 are incorrect. (5) now results in 7 being correct and 0 and 5 incorrect. And finally (2) excludes 3 being correct, so we are left with 2,6,7,8,9. Again there are some digits that can only appear in one position, namely 7 and 9: x9xx7. The first digits is not 8, the third not 6 and the fourth not 2, so we have two valid combinations: 29867 and 69287.
Due to eliminating all other possibilities these are the only correct codes.