Answer to 2)

Initial Height = 4ft
Let the constant amount be X.

After 1 year : 4+x
After 2 years : (4+x) + x = 4+2x
After 3 years : (4+2x) + x = 4+3x
.
.
.
After 6 years : 4+6x

Given,
4+6x= 4+4x + 1/5(4+4x)

Solving,X = 2/3

You must have made the mistake of writing,
4+6x=1/5(4+4x) which is wrong.

Problem 3)

Mixture 1 : 10% gravel and quantity : x tonnes
Therefore amount of gravel : 0.1x

Mixture 2 : 2% gravel,and y tonnes
Amount of gravel : 0.02y

They are mixed to produce a mixture of z tonnes that is 5% gravel

Hence, 0.1x+0.02y=0.05z
We need x.

Statement 1
-----------
y=10
Now x+y=z
x+10=z

Putting it in the equation
0.1x+10*0.02=(x+10)0.05

We can get x,hece sufficient.

Statement 2
-----------

z=16
==>x+y=16
And we have
0.1x+0.02y=0.05z
i.e
0.1x+0.02y=0.05*16

We can slve for x and y,hence sufficient,

Hence OA-D

Questions 1 and 4,i'll get back if I find the solution.Stuck myself.

Can someone solve 1st and 4th qns?? i'm not getting the required answer..

The 4th Question is a typo error. Was going through some previous posts and Stuart Kovinsky has pointed out it has to be

2^((x+y)^2)
-----------
2^((x-y)^2)

Then we use

(a^m)/(a^n) = a^(m-n) and then the expression will reduce to 2^4 = 16.