Correct answer is 3. YOU DIDN'T TRY!!!
Explanation:

• (1,2,4,7,3,6) is a 6 cycle, with order 6
• (7,5,1)(2,3) is a product of disjoint 3 and 2 cycles, so the order is lcm(3,2)=6
• (2,1,6)(2,3) is also a product of a 3 and 2 cycle, but these are not disjoint, and the product turns out to be a 4 cycle
• (1,2,3,7,5)(2,6) the product of these two cycles is a 6 cycle
• (1,3,4)(2,5,7) the order of these two disjoint 3 cycles is lcm(3,3)=3

Comment: Take care to notice whether cycles are disjoint, e.g., you might not notice this with the third and fourth examples, and think they are order 6 and 10.

Correct answer: 3 is not possible: YOU DIDN'T TRY!!!

• (1,2,3)(4,5) has order 6 = lcm(3,2)
• (1,2,3)(3,4) = (1,2,3,4) has order 4
• (1,2,3)(3,2) = (2,1) has order 2

This covers all cases, since the overlap of the two permutations is 0, 1 or 2 symbols. All cases are conjugate to one of these examples.
Comment: This is pracice with multplying out cycles.

Correct answer is C₅. YOU DIDN'T TRY!!!
For any element g, by the orbit stabiliser theorem, we have |Orb_G(g)| |Stab_G(g)| = |G| So, the order of any orbit must divide the order of the group. In the case of C₅, the order of the group is 5, so the order of an orbit is 1 or 5, but since there are only 4 elements, the order of all orbits is 1, which means the action is trivial. All the other groups have even order. For the cyclic groups, let a be the generator. Then we can make a act as (1,2)(3,4). Every time a is applied, the elements 1 and 2 switch, and the elements 3 and 4 switch. When a is applied n times where n is the order of the group, since n is even, the elements are back at their original positions, so aⁿ=1 acts trivially. This can't work if the group order is odd, but this map, aⁱ → ((1,2)(3,4))ⁱ is an action for the even order cyclic groups. For S₃, we can just map each element of S₃ to the corresponding permutation of {1,2,3}, leaving 4 fixed, so this is also a non trivial action.
Comment: This is pracice with the orbit stabiliser theorem.