已知向量a=(sinθ,1),b=(1,cosθ),-π/2＜θ＜π/2.（1）若向量a⊥b求θ,（2）求向量|a+b|的最大值

1)∵a⊥b
∴sinθ+cosθ=0
∴tanθ=-1
∵-π/2＜θ＜π/2
∴θ=-π/4
2)∵a+b=(sinθ+1,cosθ+1)
∴|a+b|
=√(a+b)²
=√[(sinθ+1)²+(cosθ+1)²]
=√[sin²θ+1+2sinθ+cos²θ+1+2cosθ+]
=√[3+2(sinθ+cosθ)]
=√[3+2√2sin(θ+π/4)]
∵-π/2＜θ＜π/2 ∴-π/4