(1)向量a·（a+2b）＝a^2+2ab=(cosα)^2+(sinα)^2+2×(cosαcosβ+sinαsinβ)＝1+2cos(α-β)
∵-1≤cos(α-β)≤1
∴-2≤2cos(α-β)≤2
∴-1≤1+2cos≤3
∴-1≤向量a·（a+2b）≤3
(2)a+2b|^2=|a|^2+4a.b+|b|^2
|a|^2=1,|b|^2=1,
4a.b=4*(cosαcosβ+sinαsinβ)=4cosα-β=2
所以|a+2b|^2=4,|a+2b|=2
(1)向量a·（a+2b）＝a^2+2ab=(cosα)^2+(sinα)^2+2×(cosαcosβ+sinαsinβ)＝1+2cos(α-β)
∵-1≤cos(α-β)≤1
∴-2≤2cos(α-β)≤2
∴-1≤1+2cos≤3
∴-1≤向量a·（a+2b）≤3
(2)a+2b|^2=|a|^2+4a.b+|b|^2
|a|^2=1,|b|^2=1,
4a.b=4*(cosαcosβ+sinαsinβ)=4cosα-β=2
所以|a+2b|^2=4,|a+2b|=2