1) We can simplify sin2a/$1+cos2a$ by using the double angle identity sin2θ = 2sinθcosθ and the Pythagorean identity sin^2θ + cos^2θ = 1.

sin2a = 2sinacos2a
1 + cos2a = 1 + cos^2a - sin^2a = 1 + cos^2a - $1-co{s}^{2}a$ = 2cos^2a

Therefore, sin2a/$1+cos2a$ = 2sinacos2a / 2cos^2a = sin2acos2a / cos^2a = sin2a/cos2a = tan2a

2) To simplify $sina+2sin(pi/3-a)$/$2sin(pi/6-a)-cosa$, we can convert the sine and cosine of π/3 and π/6 using trigonometric identities sin$\pi /3$ = √3/2, sin$\pi /6$ = 1/2, and cos$\pi /6$ = √3/2.

$sina+2sin(pi/3-a)$/$2sin(pi/6-a)-cosa$ = $sina+2sin(\pi /3-a)$/$2sin(\pi /6-a)-cosa$ = $sina+2\surd 3/2cosa$/$2(1/2)cosa-cosa$ = $sina+\surd 3cosa$/$cosa-cosa$ = $sina+\surd 3cosa$/0
= Undefined

3) $sina+cosa$^2 + $sina-cosa$^2
= sin^2a + 2sina cosa + cos^2a + sin^2a - 2sina cosa + cos^2a
= 2$si{n}^{2}a+co{s}^{2}a$ = 2$1$ = 2

4) $1-(sina+cosa{)}^2$/$sina<em>cosa-ctga$ = $1-(si{n}^{2}a+2sina</em>cosa+co{s}^{2}a)$/$sina<em>cosa-ctga$ = $1-1$/$sina</em>cosa-ctga$ = 0/$sina\ast cosa-ctga$ = 0

Therefore, the simplified forms of the given expressions are:
1) tan2a
2) Undefined
3) 2
4) 0