Step 1 :(a) (1) \(\tau^{0} = (1, 6, 4, 5, 2, 3)\) (2) \(\tau^{3} = (1, 6, 4, 5, 2, 3)\) (3) \(\sigma^{-3} = (1, 4, 3, 5, 2)\)
Step 2 :(b) (1) \(|\sigma| = 1\) (2) \(\operatorname{sgn}(\tau) = -1\) (3) \(\operatorname{sgn}(\sigma \tau) = 1\)
Step 3 :(c) \(\tau \notin A_{6}\), because its sign is -1, which means it's an odd permutation. \(\boxed{\text{Final Answer}}\)