(2)? Original formula = (2-x)/(x-1) ÷ {[(x+1) * (x-1)-3]/(x-1)}

=(2-x)/(x- 1)÷[(x^2-4)/(x- 1)]

=(2-x)/(x- 1)÷{[(x+2)*(x-2)]/(x- 1)}

=[(2-x)/(x- 1)]×{(x- 1)/[(x+2)*(x-2)]

=- 1/(x+2)

3? :( 1)? ① When t≤3, y = 0.2.

② when t > 3 (t (score) table is not a positive integer), y = 0.2+0.1t (integer part of t = 1).

(2) See figure

(3)① When 0 < t ≤ 3, the telephone charges before and after adjustment are the same in 0.2 yuan.

2 When? 3 < t ≤ 4, 0.4 yuan before adjustment and 0.3 yuan after adjustment.

3: When? When 4 < t ≤ 5, the telephone charges before and after adjustment are the same in 0.4 yuan.

4: When? 5 < t ≤ 6, 0.4 yuan before adjustment and 0.5 yuan after adjustment.

Therefore, when 0 < t ≤ 3 and 4 < t ≤ 5, the telephone charges before and after adjustment are the same.