Is it A+B/2 or (A+B)/2?

If it is (A+B)/2, the following is the solution:

Judging from the meaning of the question:

4sin2[(A+B)/2]-cos2C=7/2

= > 2-[2-4 sin 2[(A+B)/2]]-cos2C = 7/2

= > 2-2[ 1-2 sin 2[(A+B)/2]]-cos2C = 7/2

= > 2-2 cos(A+B)-(2 cos 2c- 1)= 7/2

Because a+b+c = л

= > 2-2COS(л-C)-2COS 2c+ 1 = 7/2

= > 2+2 cos-2 cos 2c- 1 = 7/2

=> 4COS2C-4cosC+ 1=0

=> cosC= 1/2

=> C=60？

As for the ABC area, I don't remember it clearly because I learned it six years ago. There are formulas such as a/sinA=b/sinB, and then the triangle area seems to be S= 1/2*a*b*sinC and so on. Look up the formula in the book and try to push it ~