arc AB = 122 degrees
angle ADB = 61 degrees
Arc BC is 58 degrees. The arc AB combines with it to form semicircle ABC, which is basically arc ABC with measure 180
This must mean arc AB is 180-58 =122 degrees
When in doubt, chances are the number 180 might be involved (in some way) when it comes to geometry problems. Of course, this isn't always a guarantee. The number does show up a lot though.
Once we determine what arc AB is equal to, we cut it in half to get the inscribed angle ADB. Refer to the inscribed angle theorem. Notice how angle ADB subtends the arc AB.
x = length of AE
Since AC = 13, this makes EC = 13-x
We'll use the intersecting chords theorem to find x.
(AE)*(EC) = (BE)*(DE)
x*(13-x) = 6*6
13x-x^2 = 36
-x^2 + 13x - 36 = 0
x^2 - 13x + 36 = 0
(x-9)(x-4) = 0
x-9 = 0 or x-4 = 0
x = 9 or x = 4
This means AE could be 9 units long or 4 units long.
Usually the diagram isn't drawn to scale so we don't really use it all too much (other than initial set up). However, if the diagram was drawn to scale, then notice that AE is over the halfway point of AC. This means we would be able to rule out x = 4 since 4 is not over the halfway point of 13. I have a feeling this is what your teacher might be going for.
In short, only AE = 9 is possible.