[Solution]
∫∠aeb = 90°, ∴AB is the diameter of the circumscribed circle of △ABE, and the midpoint of ∴AB is the center of the circle.
From the midpoint coordinate formula, it is easy to find that the center coordinate is (-2,0).
∵| AB | = 3- 1 = 2, and the radius of the circle of∴ is 1.
Rewrite the straight line y = x+t into a general formula and get: x-y=x+t = 0.
The ∴ distance from the center of the circle to the straight line x-y+t = 0 = |-2-0+t |/√ (1+1) = | 2-t |/√ 2.
∵ There are two points E on the straight line X-Y+T = 0, so∠ ∠aeb = 90°, and∴ Circle intersects with the straight line X-Y+T = 0.
∴ The distance from the center of the circle to the straight line x-y+t = 0 is less than the radius of the circle, ∴| 2-t |/√ 2 < 1, ∴| 2-t | < √ 2.
1. When 2-t≥0, there are 2-t < √ 2, ∴ t > 2-√ 2.
From 2-t≥0, we get t≤2.
The trade-off range of ∴, t at this time is (2-√ 2,2].
2. When 2-t < 0, there are: t-2 < √ 2, ∴ t < 2+√ 2.
And from 2-t < 0, t > 2.
At this time, the trade-off range of ∴ and t is (2,2+√ 2).
To sum up, the acceptable range of t is (2-√ 2,2+√ 2).
Note: Please check the original question carefully. If the original question is not my guess, please supplement it.