I'm not sure what you mean. In any event I believe the answer is incorrect - according to my worked analysis.
It usually works the other way on the homework forum. You show your working first, then I may show you mine or (better still) offer some suggestions on how you might go about finding the solution.
really i founded the solution under the attached question But u told me it's incorrect According to my information i know when the mutual inductance (M) is minus, added M to L1 and L2
If you will show how you derived your answer then we could help you correct any errors you may have made.
Consider the original circuit. If you apply a "positive" AC voltage to the V1 terminals, you get a "positive" voltage at the V2 terminals (no phase inversion, in other words). But, your equivalent has a shunt element of -2H. Since the voltage that appears at the open-circuited V2 terminals is just the voltage across that shunt element, you will get phase inversion due to the minus sign, so for that reason alone your equivalent is not right. Re-examine how you formed your equations.
Your second equation should be -V2 = 5 s I2 - 2 s I1 Try reversing the direction of I2 and see what equations you get, both for the original transformer and for the tee equivalent with 3H, 2H and 3H.