1)当x＜0时,f(x)=0;
当x≥0时,f(x)=2x-1/2^|x| .
由条件可知2x-1/2^|x|=2,即22x-2·2x-1=0,
解得2x=1±√2.
∵2x＞0,∴x=log2(1+√2).
(2)当t∈〔1,2〕时,2^t*(2^2t-1/2^2t)+m(2^t-1/2^t)≥0,10分
即m(2^2t-1)≥-(2^4t-1).
∵2^2t-1＞0,∴m≥-(2^2t+1).
∵t∈〔1,2〕,∴-(1+2^2t)∈〔-17,-5〕,
故m的取值范围是〔-5,+∞).