1。由题意得。fx的递增区间取决于2sin(2x+π/6)的区间,且与sin函数的增减区间相同所以2kπ-π/2<2x+π/6<2kπ+π/2,kπ-π/32kπ+π/2<2x+π/6<2kπ+3π/2，kπ+π/62。在x∈[0，π/2]时，(0,π/6)上递增，在(π/6.π/2)上递减，最大值在π/6点上fx=2sin(2x+π/6)+a+1=2+a+1=4,所以a=1fx=2sin(2x+π/6)+23.在2sin(2x+π/6)最大时，f(x)最大，所以有 2x+π/6=2kπ+π/2 x=kπ+π/6
