I'll work on the 50% solution for you and you can do the 30% solution.
You have 1 litre of a 70% acid solution so the solution contains 0.7 litres of acid. You want to add enough water so that the 0.7 litres of acid accounts for 50% of the solution. To say this algebraically I need a variable for the amount of water I add so let x be the number of litres of water I add. So now I have
1 + x litres of solution which contains 0.7 litres of acid.
Thus, if this 0.7 litres of acid is 50% of the solution then
0.7 = 0.5 (1 + x)
You can now solve this algebraically for x but if you look at this equation it says
0.7 is half of 1 + x
Hence 1 + x must be 1.4 and thus x = 0.4 litres.
Now you try the 30% solution.