What is the exact value of cot(arccos(1/sqrt5))?

We'll assume an acute answer here. The expression arccos(1/sqrt5)) means "the angle whose cosine is 1/sqrt5. So draw a right triangle. Mark one of the acute angles as x, and label the hypotenuse sqrt5, and the side adjacent to angle x as 1. Now use Pythagoras to find the length of the remaining side, and you will be able to find the cotangent of x.