(6) + (7): excludes 2 due to (7) having only correctly placed digits and (6) only incorrectly placed ones.
(5) + (8): (5) has two correct out of 65917, (8) has three correct out of 65198. The only change of 7 -> 8 adds one correct digit, thus 8 is included, 7 is excluded.
(7) fixes the found 8 to x8xxx
(5) + (6): out of 6591 two are correct, and out of 456 one is correct, thus at most one of 56 is correct.
two possibilities:
1. none out of 56 correct: 4 has to be correct, 91 have to be correct. (7) implies that 0 is correct, thus x8xx0. (5) has to have one correctly placed digit. Since the last digit is already fixed it has to be x8x90. With 4 and 1 other digits this cannot be a multiple of 11. (4-8+1-9+0=-12)
2. one out of 56 is correct. As a result 4 is excluded. Now one out of 56 is true and one out of 91 is true. We have excluded 2,4,7
If 5 is true, 6 is excluded. Also (7) fixes it to x85xx and excludes 0.
Now (5) requires the one out of 91 that is true to be correctly placed.
If 9 is true, 1 is excluded and the code is a859b with 0,1,2,4,6,7 excluded. For it to be a multiple of 11 a-8+5-9+b = a+b-12 must be a multiple of 11. This is fulfilled by a+b=1 and a+b=12. a+b=1 can not be fulfilled. a+b=12 is fulfilled by 3 and 9. 4 and 8, 5 and 7, 6 and 6 all are excluded. The hints do not offer further constraints for the digits 3 and 9, thus not leading to a unique solution.
If 1 is true, 9 is excluded and the code is a85b1 with 0,2,4,6,7,9 excluded. For it to be a multiple of 11, a-b=2 has to hold. There are the two possibilities a=3, b=1 and x=a, y=b, again not offering a unique solution.
Thus 5 is excluded, 6 is correct. (7) then requires the code to look like x8xx0. 2,4,5,7 are excluded.
If 9 is true, then (5) forces either 689a0, which is excluded, since for it to be a multiple of 11 a has to be 5, or a8690, which contradicts (6).
Thus 9 is excluded and 1 is true. (5) forces the code to be 68ab0, with a-b=2 and one of these being 1. Thus only 68310 remains.