Get cosC+cos(A-B)= 1-cos2C, cos(A-B)-cos(A+B)=2sin2C.

SinAsinB=sin2C, according to sine theorem, ab=c2, ①,

From sine theorem and (b+a)(sinB-sinA)=asinB, b2-a2=ab, ②, from ① and ②, b2=a2+c2,

So △ABC is a right triangle with b = 90;

(ⅱ)∫a+c = 90 ,∴sin2c=sinasinb=sina=cosc，

So cos2C+cosC- 1=0, and the solution is cosC=

- 1+5

2

Or cosC=

- 1-5

2

(give up),

That is, cosC=

- 1+5

2 .