Finding the concentration from a dilution
Solving one out of the four would be good enough to understand the concept behind this. Therefore, I shall explain the first experiment and the concentrations of the components in the solution produced.
In experiment 1 all four are added in equal quantities. Finding the total volume, we get the total volume
$$V_\text{tot} = \pu{4 ml}.$$
Assuming the amount of $i$th solute to be denoted by $n_i,$ its molarity $c_i$ can be found as
$$c_i = \frac{n_i}{V_\mathrm{tot}}.$$
Now, we add $\pu{1 ml}$ of $\pu{2.0 M}$ acetone. This is equivalent to saying we add $1/1000$ of $\pu{2 mol}$ of acetone. Therefore we add $\pu{2 mmol}$ (millimoles) of acetone.
This means that we have $\pu{2 mmol}$ of acetone in $\pu{4 mL}$ of water. Using the formula for molarity we get that acetone is in $\pu{0.5 M}$ in solution.
Hope you can follow on for the other parts as well, the concept remains the same.