Will someone please help me with this? Thank you so much!!!

Well, let's go back through #1 - 5# then. You'll have to be careful with sig figs... I've given myself one more sig fig because #"0.1 M"# is just sad. I've seen two sig figs very often on concentrations.

I got #4.9%#.

The mols of #"NaOH"# are found simply by the volume of it added to acetic acid.

#"0.10 mol NaOH"/cancel"L" xx 74.1 cancel"mL" xx cancel"1 L"/(1000 cancel"mL") = ul"0.0074(1) mols"#

And they react in a #1:1# ratio, so this is also the mols of acetic acid.

Therefore, in the #"10 mL"# of vinegar, #"0.0074(1) mols"# of it was acetic acid that reacted, resulting in a molarity for trial 1 of

#"0.0074(1) mols acetic acid"/"0.010 L" = ul"0.74(1) M"#

and in trial 2 a molarity of

#"0.0090(2) mols acetic acid"/"0.010 L" = ul"0.90(2) M"#

for an average molarity of

#("0.74(1) M" + "0.90(2) M")/2 = ul"0.82(2) M"#

So, the grams of acetic acid per liter is just the mass of it in #"1 L"#:

#(0.82(2) cancel"mols acetic acid")/"L" xx ("60.05 g CH"_3"COOH")/cancel("1 mol") = ul"49.(3) g/L"#

And so, the mass percent in vinegar is the mass of acetic acid in #"1 L"# divided by the mass of #"1 L"# of vinegar.

The mass of #bb"1 L"# of vinegar is:

#cancel"1 L" xx "1.005 g"/cancel"mL" xx (1000 cancel"mL")/cancel"L" = ul"1005 g vinegar"#

And so, the mass percent is:

#color(blue)(%"w/w") = "49.(3) g acetic acid"/"1005 g vinegar" xx 100%#

#= color(blue)ul(4.9%)#