*** The volume of the solutions, which is missing, is required to figure out the answer.

*** I assume that 1L of each solution is needed.

For acetic acid:

First determine the dilution factor. For 0.08M solution: 2.00 M/ 0.08M = 25. Thus 2M solution needs to be diluted 25 times. For 1L of the diluted solution we will need 1000 mL / 25 = 40 mL

For 1L of 0.160 M solution we could use analogous reasoning or simplify a little by noting that the second solution is twice as concentrated as the first, thus requiring twice as much of 2M solution or 80 mL

For Na2CO3:

First, find the molar mass of Na2CO3: 2*23 g/mol +12 g/mol +3*16 g/mol = 106 g/mol

Second, find the quantities needed by converting the molar concentrations into the weight in grams in the solutions.

1 L of 0.1 M solution contains 106 g/mol * 0.1 mol/L = 10.6 g

1 L of 0.1 M solution contains 106 g/mol * 0.14 mol/L = 14.84 g