Calculating mass and volume of two different substances required for one solution
We have direct formula for calculating the weight of the substance as,
$$\text{Weight}=\frac{\text{Molarity} \times \text {molecular wt} \times \text{volume in mL}}{1000}$$
$$=\frac{0.02\times 250 \times 100}{1000}$$
$$=0.5 \text{ g}$$

So to prepare 0.02 $ \mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ ,0.5 g of copper(II) sulfate pentahydrate should be dissolved in 100 mL water.

To find the volume of ammonia,

Initial molarity = 2 M       

Initial volume= ?

Final molarity = 0.2 M       

Final volume = 100 mL

We have formula, 
 $$ \text {Initial molarity} \times \text{Initial volume} = \text {Final molarity} \times \text{Final volume}$$
$$\text{Initial volume} = \frac{0.2 \times 100}{2}$$
$$= 10 mL$$

Hence, 10 mL of 2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$ should be added in 100 mL water to obtain 0.2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$.