= 1-3√2(√2/2cosx+√2/2sinx)

= 1-3√2sin(x+π/4)

(1) Let 2kπ+π/2 ≤ x+π/4 ≤ 2kπ+3π/2k∈ n.

2kπ+π/4≤x≤2kπ+5π/4 k∈N

The monotonic increasing range is [2kπ+π/4, 2kπ+5π/4] (k∈N).

(2)f(x)= 1-3√2 sin(x+π/4)=- 1

sin(x+π/4)=√2/3

X∈(π/2, 3π/4) then 3π/4 ≤ X+π/4 ≤

cos(x+π/4)=-√7/3

tan(x+π/4)=-√2/√7

tan2x =-cot(2x+π/2)=- 1/tan 2(x+π/4)=[tan？ (x+π/4)- 1]/2tan(x+π/4)= 5√ 14/28