A wave pulse travels down a slinky. The mass of the slinky is m = 0.86 kg and is initially stretched to a length L = 7.8 m. The wave pulse has an amplitude of A = 0.26 m and takes t = 0.472 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.43 Hz.
Now the slinky is stretched to twice its length (but the total mass does not change). (assume the slinky acts as a spring that obeys Hooke's Law)

What is the new time it takes for a wave pulse to travel down the slinky? I am saying this would be the same as the original time, and the answer would be 0.472s, but I am not sure about this...

When you stretch the slinky, two different things happen at the same time to change the propagation velocity: the tension in the slinky is doubling (thereby increasing the velocity by a factor of sqrt2) and the mass per unit length cut in half (thereby also increasing the velocity by a factor of sqrt2). Thus, combining both effects, the velocity is doubled. Offsetting this effect is the fact that the distance the wave must travel also doubles. Hence, the new time should indeed be the same as the original time. I didn't check your math, but I'm pretty sure your answer is correct.