Solution:

(1) Let the subtraction be100a+10b+c)-(100c+10b+a) = 99 (a-c) = X.

X must be a multiple of 99. A-c= 1, 2, 3 ... can all be satisfied.

(2) The middle of addition must be 9.

Make it possible to add

( 100 a+ 10 b+c)+( 100 c+ 10 b+a)= 1089

10 1(a+b)+20b = 1089

If 1089 is divisible by 10 1, then b must be 9... 1.

For example: 963-369=594 594+495= 1089.

45 1- 154=297 297+729= 1089

67 1- 176=495 495+594= 1089