因为函数f(x) = 3sin(2x-π/4)
(1)T=2π/2=π
(2)因为x∈[π/6,3π/4]
所以：2x-π/4∈[π/12,5π/4]
所以sin(2x-π/4)∈[-√2/2,1]
所以3sin(2x-π/4)∈[-3√2/2,3]
即函数值域为：[-3√2/2,3]