已知函数f（x）=2sin²x+2倍根号3sinxcosx-1,

1
∵2sinxcosx=sin2x
2sin²x-1=-cos2x
∴f(x)=2sin²x+2√3sinxcosx-1
=√3sin2x-cos2x
=2(√3/2sin2x-1/2cos2x)
=2sin(2x-π/6)
当2x-π/6=2kπ-π/2,k∈Z
即x=kπ-π/6,k∈Z时,
f(x)取得最小值-2
此时x集合{x|x=kπ-π/6,k∈Z}
(2)
当 x∈[π/6,π/2]时,
2x∈[π/3,π]
2x-π/6∈[π/6,5π/6]
∴1/2≤sin(2x-π/6)≤1
∴1≤f(x)≤2
∵方程f(x)=log2 (t-1) 在 x∈[π/6,π/2]上有实数解
∴1≤log₂(t-1)≤2
∴2≤t-1≤4
∴3≤t≤5

希望能帮助你！