(1) when x =112π, it is the function f (x) = 2sin (2×112π-π 3) =-2.
At this time, the function takes the minimum value, so x =112π is a symmetry of the image of the function; Correct.
② Because 2kπ-π2≤2x-π3≤2kπ+π2 k∈Z, when solving kπ-π 12≤x≤kπ+5π 12, the function monotonically increases.
Let k=0, -π 12≤x≤5π 12, and the function is increasing function (-π 12, 5π12) in the interval; So 2 is correct.
③ When x=23π, the function f(x)=2sin(2×23π-π3)=0,
Image c is symmetric about point (2 π 3,0); correct
④ The image of function f(x) can be shifted π3 to the right from the image of function y=2sin2x to get f (x) = 2sin (2x-2sin3x-π 3) =-2sin2x, which is incorrect.
So choose a.