（Ⅰ）由题意可得 [2π/ω]=π，∴ω=2．
∵图象的一条对称轴是直线x＝
π
8，
∴2cos（2×[π/8]+φ）=±2．
再由，-π＜φ＜0可得 φ=-[π/4]，
∴函数f（x）=2cos（2x-[π/4]）．
（Ⅱ）令 2kπ≤2x-[π/4]≤2kπ+π，k∈z，解得 kπ+[π/8]≤x≤kπ+[5π/8]，k∈z，故函数的减区间为[kπ+[π/8]，kπ+[5π/8]]，k∈z．
（Ⅲ）列表：
x 0 [π/8] [2π/8] [3π/8] [4π/8] [5π/8] [6π/8] [7π/8] π
2x-[π/4] -[π/4] 0 [π/4] [π/2] [3π/4] π [5π/4] [3π/2] [π/4]
f（x）