Injective: let (2x- 1)/4 = (2y- 1)/4, and x=y is obtained, so it is injective.

Surjective: Obviously, the range of f(x) is R, so it is surjective.

Question (2)

Let f(y)=x, that is, (2y- 1)/4=x, and the solution is y=2x+ 1/2.

So f(x)=2x+ 1/2.

(f？ g)(x)= f(g(x))= f(x+ 1)= 2(x+ 1)+ 1/2 = 2x+5/2