Let t= sinx+cosx=√2sin(x+∏/4), ∴t∈[-√2, √2].

Then, T2 = (sinx+cosx) 2 =1+2 sinx cosx.

Then sinx cosx = (t 2- 1)/2.

∴y=t+(t^2- 1)/2=( 1/2)t^2+t-( 1/2)

Let y = g (t) = (1/2) t2+t-(1/2), t ∈ [-√ 2, √ 2].

The symmetry axis is t=- 1, and the opening is upward.

∴ Maximum value y = g (√ 2) = (1/2) (√ 2) 2+√ 2-(1/2)

=√2+ 1/2

The minimum value y = g (-1) = (1/2) (-1) 2+(-1)-(1/2).

=- 1

So the range is [- 1, √2+ 1/2].

(2)y = cos(2x/5)+sin(2x/5)=√2 sin[(2x/5)+∏/4]

The minimum positive period is T=2∏/(2/5)=5∏.

The distance between two adjacent symmetry axes is d= T/2=5∏/2.

(3)∵4tan(a/2)= 1-〔tan(a/2)〕^2

∴2tan(a/2)/{ 1-[ tan (a/2)] 2} =1/2.

∴tana=2tan(a/2)/{ 1-[ tan (a/2)] 2} =1/2.

∵ 0

∴sina=√5/5,cosa=2√5/5

∴sin2a=2sinacosa=4/5,cos2a=3/5

∫3 sinb = sin(2a+b)= sin 2 acosb+sinb cos 2 a。

3sinb=(4/5)cosb+(3/5)sinb

Simplify and acquire

12sinb=4cosb

∴tanb=sinb/cosb= 1/3

∴tan(a+b)=(tana+tanb)/( 1-tanatanb)

=[( 1/2)+( 1/3)]/[ 1-( 1/6)]

= 1

∴a+b=∏/4