解1由f（x）=4coswx×sin（wx+∏/4）
=2×[sin（（wx+π/4)+wx）+sin（（wx+π/4)-wx）]
=2sin（2wx+π/4）+2sin（π/4)
=2sin（2wx+π/4）+√2
故T=2π/2w=π/w
又由T=π
即π/w=π
即w=1
（2）由f（x）=2sin（2x+π/4）+√2
由x属于[0,π/2]
则2x属于[0,π]
即2x+π/4属于[π/4,5π/4]
即2x+π/4属于[π/4,π/2],即x属于[0,π/8]时,f（x）=2sin（2x+π/4）+√2是增函数
2x+π/4属于[π/2,5π/4],即x属于[π/8,π/2]时,f（x）=2sin（2x+π/4）+√2是减函数.