More Projectile motion equations!

OK, so I have to solve some equations, which I've done but I'm not sure they are right. Can somebody please go through my work and tell me if it is wrong or right. If its wrong can you please point out the mistake?

A ball is thrown upwards with an initial velocity of 5m/s from posiiton A on a building with a height of 13.2m. The ball goes to a maximum height at B and then falls down alongside the building to the ground.

Q1: Calculate the maximum height taht the ball will reach above the ground.
A: Take up as -
Vf^2 = Vi^2 + 2a Δx
0 = (-5)^2 + 2(9.8)Δx
Δx = -1.28

.'. Δx = 1.28 + 13.2
= 14.48 m

Q2: Calculate the velocity with which the ball hit the ground"
A: Vf^2 = Vi^2 + 2a Δx
= (-5)^2 + 2(9.8)(14,48 x 2)
= 25 + 567.62
Vf = 24, 34 m/s

Q3: Calculate the time it will take the ball to reach maximum height.
A: Δx = ((Vi + Vf) / 2) Δt
14.48 = (-5/2) Δt
Δt = -5. 79

.'. Δt= 5.79 s

Q4:Calculate the time it will take the ball from point A to the ground.
A: Δx = 1.28 + 1.28 + 13.2
= 15.76

.'. Δx = ((Vi + Vf) / 2) Δt
15,76 = (-5/2) Δt
Δt = -6.30

.'. Δt= 6.30s

Q5: Find the positon of the ball after 1.5s
A: Δx = ((Vi + Vf) / 2) Δt
= (-5/2) (1.5)
= -3.75
.'. Δx = 3.75 upwards

o.k your answer 1 is correct

however others are not correct.

2) final velocity V = square root of ( 5^2 + 2 x9.8 x 13.2) = 16.84 m/sec

3) time taken to reach maximum height t = u / g = 5 /9.8 = 0.51 sec

4) time taken to reach ground from A is t = u + square root of (u^2 + 2gh ) / g = 2.23 sec

5) use s = ut + 1/2 at^2, to find its position after 1.5 sec.

Q2 is done wrongly you have to take motion here from highest point where ball stops to ground,u can't take it from building to up and then down,because during upward motion a=-9.8m/s^2,and positive for downward motion,
v^2=u^2+2ax
= 0+2*9.8*14.8