You have 12 choices for the first student chosen, 11 choices for the next, then 10, then 9. However, this over-counts everything by a factor of 4! (the number of ways in which four objects can be arranged with regard to order).
Thus, the answer is
$$\frac{12\cdot 11 \cdot 10 \cdot 9}{4!} = 495$$
Method 1: Suppose we have four green balls, two of which are placed in a box, and six blue balls. Line up the six blue balls in a row. This creates seven spaces, five between successive blue balls and two at the ends of the row. Choose one of these seven spaces for the box with two green balls. Choose two of the remaining six spaces for the other two green balls. Now number the balls from left to right. The numbers on the green balls represent the positions of the selected people. Notice that exactly two of the green balls are consecutive, namely the ones in the box. Hence, there are
$$\binom{7}{1}\binom{6}{2} = 105$$
ways to select four people from a row of $10$ people so that exactly two of the adjacent people are consecutive.
Observe that this is essentially your second method condensed into a single step. There are five ways to place the box with two green balls between two blue balls and two ways to place the box at an end of the row of six blue balls. This accounts for the factor of $7$. In each case, we are left with six spaces, two of which must be filled with the remaining green balls.
Method 2: In your first approach, you overlooked the possibilities that no two selected people are adjacent and that two separated disjoint pairs of adjacent people are selected. There are $\binom{10}{4}$ ways to select four of the ten people. From these, we must exclude the four cases below.
No two selected people are adjacent: We line up six blue balls in a row, creating seven spaces. We choose four of these seven spaces in which to place a green ball, then number the balls from left to right. The numbers on the green balls represent the positions of the selected people. There are $\binom{7}{4}$ such cases.
Two separated disjoint pairs of adjacent people are selected: We line up six blue balls in a row, creating seven spaces. We choose two of the seven spaces in which to place two boxes, each of which contains two green balls, then numbers the balls from left to right. The numbers on the green balls represent the positions of the selected people. There are $\binom{7}{2}$ such cases.
A block of three consecutive people and a fourth person not adjacent to them are selected: We line up six blue balls in a row, creating seven spaces. We choose one of the seven spaces for a box with three green balls and one of the remaining six spaces for the other green ball, then number the balls from left to right. The numbers on the green balls represent the positions of the selected people. There are $\binom{7}{1}\binom{6}{1}$ such cases.
A block of four consecutive people is selected: We line up six blue balls in a row, creating seven spaces. We choose one of the seven spaces for a box with four green balls, then number the balls from left to right. The numbers on the green balls represent the positions of the selected people. There are $\binom{7}{1}$ such cases.
Thus, the number of permissible selections is
$$\binom{10}{4} - \binom{7}{4} - \binom{7}{2} - \binom{7}{1}\binom{6}{1} - \binom{7}{1} = 105$$
Best Answer
For illustration, imagine the same problem but with a five element set: {A,B,C,D,E}.
Part 1: As @lulu mentioned, there are two choices that determine the other two choices. Therefore, you can have: AB,AC,BC,AD,BD,CD,AE,BE,CE,DE as the pairs of two choices. If order mattered, then we would be able to count the 4-sequences for each pairing, ex: AABB,ABAB,ABBA,BABA,BAAB,BBAA for pairing AB. Since order doesn't matter, these are all equivalent to the subset {A,A,B,B}. Based on the question's use of "chosen", I am left to assume the sequence does not matter, only the set. In this case, there are ${5 \choose 2 } = 10$ possible sets, as we enumerated above.
Part 2: Since we are choosing sets, sequences do not matter. In this choice, we still choose two elements that determine the set. If order mattered, we would have sequences like AAAB,AABA,ABAA,BAAA for choice AB. Since order doesn't matter, these are all equivalent to the subset {A,A,A,B} and the answer is still the same: ${5 \choose 2 } = 10$ possible sets. However, as pointed out in the comment below, {B,B,B,A} is the other possible subset, so we multiply by two to arrive at 20 possible sets.
Therefore, from the {A,B,C,D,E,F,G,H,I}, your answers are ${9 \choose 2 } = 36$ possible sets for part 1 and ${9 \choose 2 }\times 2=72$ for part 2, and the unique parings come from two choices: AB,AC,BC,AD,BD,CD,AE,BE,CE,DE,AF,BF,CF,DF,EF,AG,BG,CG,DG,EG,FG,AH,BH,CH,DH,EH,FH,GH,AI,BI,CI,DI,EI,FI,GI,HI.
Bonus: if order did matter, and we are now talking about sequences:
Part 1: Use the multiplication principle. We have four locations, and choose two, the other two are determined from the constraint: ${4 \choose 2 } = 6$. Then, to fill the two chosen locations, we may choose ${5 \choose 2 } = 10$ different pairs. Therefore, the number of sequences is ${4 \choose 2 } {5 \choose 2 } = 6\times 10 = 60$, which can be written out and verified.
Part 2: Same idea, except now we choose 3 or equivalently, 1: ${4 \choose 3 } = {4 \choose 1 } = 4$ and the number of sequences is ${4 \choose 3 } {5 \choose 2 } = 4\times 10 = 40$. As in the choice problem, since we could have {A,A,A,B} or {B,B,B,A} from choice AB, we multiply by two again, to arrive at 80 sequences.
Therefore, for your question: