Calculating the value of x in the compound CuSO4xH20 . The compound is hydrous copper sulphate crystals .?

Molar mass of CuSO₄ = (63.5 + 32.1 + 16.0*4) g/mol = 159.6 g/mol
Molar mass of H₂O = (1.0*2 + 16.0) g/mol = 18.0 g/mol

In the table, the masses of CuSO₄ and that of H₂O are wrongly calculated.
Mass of CuSO₄ = (Mass of Test Tube with CuSO₄) - (Mass of Test Tube)
Mass of H₂O = (Mass of Test Tube with CuSO₄•xH₂O) - (Mass of Test Tube with CuSO₄)

The table with corrected data is shown below.

The total mass of CuSO₄ = (0.65 + 0.68 + 0.60) g = 1.93 g
and the total mass of H₂O = (0.35 + 0.32 + 0.40) g = 1.07 g

Mole ratio CuSO₄ : H₂O
1.93/159.6 : 1.07/18.0 = 1 : x
(1.93/159.6) x = 1.07/18.0
x = (1.07/18.0) * (159.6/1.93)
x = 4.92
x ≈ 5

(You may also use average masses of CuSO₄ and H₂O instead of total masses in the calculations.)

Molar mass of CuSO4 is 159.60 g/mol
Molar mass H2O = 18.0g/mol
You ask: How do I use stoichmetry to find the number of atoms in the chemical formula ?
You should ask: How do I find the number of moles water in 1 mol CuSO4.nH2O. In other words , solve for n .

If you evaporate the water from 3 X 1 g samples of hydrated CuSO4.nH2O , you should expect the mass of the anhydrous CuSO4 to be the same for all 3 samples. You do not get this. There is some error in your work.
Let us concentrate on sample 1,
Mass of anhydrous CuSO4 = 0.35g
Mol anhydrous CuSO4 = 0.35/159.60 = 0.00219 mol
Mass of H2O = 0.65g
Mol H2O = 0.65/18.0 = 0.0361 mol
Ratio : mol CuSO4:H2O = 0.00219 : 0.0361
Ratio = 0.0361/0.00219 = 1:16.5
This is how you do the calculations for this problem. The correct answer is CuSO4.5H2O. Your experiment contains errors

Edit later : I have rechecked you submission, which I took at face value.
The main problem is that you have reversed the mass of CuSO4 and H2O
Check:
Mass of test tube + CuSO4 + test tube = 11.85g
Mass of test tube = 11.20g
Mass of CuSO4 = 11.85 - 11.20 = 0.65g
And mass of H2O = 0.35g
Now repeat the calculations
Mol CuSO4 in 0.65g = 0.65/159.6 = 0.00407mol
mol H2O in 0.35g H2O = 0.35/18 = 0.0194mol
Ratio = 0.0194 / 0.00407 = 4.77
Which is very close to the required 5mol H2O in 1 mol CuSO4.5H2O
I have often seen this experiment carried out . Invariably the CuSO4 is not fully dehydrated so as to remove all the water. This is what has happened here.

For trial 2:
Mass of CuSO4 = 15.60 - 14.92 = 0.68g
Mass of H2O = 0.32g
n = 4.17

For trial 3 figures correct
n = 5.9

If you average these 3 results you get a good result: n = 4.95