设函数y=√2sin(2ax+4π)的最小正周期为2π/3.

(1) y=√2sin(2ax+4π)= √2sin(2ax).
a>0时,最小正周期为2π/(2a)= 2π/3,a=3/2.
(2) y=f(x)= √2sin(3x).
y=f(x)的图像向右平移π/2个单位长度可得：
y=√2sin(3(x-π/2))= √2sin(3x-3π/2)= √2cos(3x).
2kπ≤3x≤2kπ+π,k∈Z.==》2kπ/3≤x≤2kπ/3+π/3,k∈Z.
==》y=g(x)的单调递减区间为[2kπ/3,2kπ/3+π/3] ,k∈Z.
2kπ-π≤3x≤2kπ,k∈Z.==》2kπ/3-π/3≤x≤2kπ/3,k∈Z.
==》y=g(x)的单调递增区间为[2kπ/3-π/3,2kπ/3] ,k∈Z.

(1) y=√2sin(2ax+4π)= √2sin(2a)
最小正周期为2π/(2a)= 2π/3, a=3/2.