(1)因為,x∈［0,π/2］,
2x＋π/6∈［π/6,7π/6］,
sin(2x＋π/6)∈［-1/2,1］,
又 a＞0
所以,-2a＋2a＋b=-5
a＋2a＋b=1
解得：a=2,b=-5
(2) 由(1)知,f(x)＝-4sin(2x＋π/6)-1
由題意 g(x)＝f(x＋π/2)
＝-4sin(2x+π+π/6)-1
=4sin(2x＋π/6)-1＞1
即sin(2x＋π/6)＞1/2
所以2x＋π/6∈(2kπ+π/6,2kπ+5π/6)
單調(diào)增區(qū)間滿足 2x＋π/6∈(2kπ+π/6,2kπ+π/2]
單調(diào)減區(qū)間滿足 2x＋π/6∈[2kπ+π/2,2kπ+5π/6)
解得 g(x)的單調(diào)增區(qū)間為 (kπ,kπ+π/6]
單調(diào)減區(qū)間為 [kπ+π/6,kπ+π/3]
打字不易,如滿意,望采納.第4行，當x=6/派的時候，y不等于1呀對不起，打錯了當x=0的時候，sin(2x+6/派）不等于0不等于1。。