Remainder problems in DS often require TESTING CASES:

If the positive integer n is greater than 6, what is the remainder when n is divided by 6?

We don't have much given information to unpack here, so let's move on to the statements:

1) When n is divided by 9, the remainder is 2.

Translate: n = (some multiple of 9) + 2.

Case 1: n = 11
when n is divided by 6, the remainder is 5.

Case 2: n = 20
when n is divided by 6, the remainder is 2.

Since we get 2 different answers in 2 different cases, this is insufficient.

2) When n is divided by 4, the remainder is 1.

Translate: n = (some multiple of 4) + 1.

Case 1: n = 9
when n is divided by 6, the remainder is 3.

Case 2: n = 13
when n is divided by 6, the remainder is 1.

Since we get 2 different answers in 2 different cases, this is insufficient.

(1) & (2) together

n must be most 2 more than a multiple of 9 and 1 more than a multiple of 4. To find cases that fit this criteria, list out numbers that are 9x + 2, then find the ones that are 1 more than a multiple of 4:
11
20
29
38
47
56
65
74
83
92
101

For each of these, test the remainder when divided by 6:

If n= 29, then the remainder when n is divided by 6 = 5.
If n= 65, then the remainder when n is divided by 6 = 5.
If n= 101, then the remainder when n is divided by 6 = 5.

Clearly there is a pattern: we will always get a remainder of 5.

The answer is C.