This a problem in physical chemistry, not a mathematics problem, although some calculations are involved.
The question provides no information on equilibrium constants, so in order to help provide you with a solution, I need to research the chemistry about pH and titration.
I believe I can answer the first part if I assume that acetic acid has an equilibrium constant K of about 1.85×10⁻⁵. Concentration of acetic acid is 0.1 mole.
We need an ICE table and an equilibrium equation (aq=aqueous solution):
CH₃COOH(aq) + H₂O(liquid) ⇌CH₃COO⁻(aq) + H₃O⁺(aq)
INITIAL 0.1 0 0 0
CHANGE -x 0 x x
EQUILIBRIUM 0.1-x 0 x x
where x is the charge transferred from acetic acid.
K=(x)(x)/(0.1-x)=x²/(0.1-x).
This produces the quadratic equation:
K(0.1-x)=x², x²+Kx-0.1K=0.
(Another way to solve this is to assume x will be very small compared to 0.1, so the equation simplifies into x²=0.1K, x=√0.1K, so x=√1.85×10⁻⁶=0.00136 approx.)
Using the quadratic formula:
x=(-K±√(K²+0.4K))/2, x=(-K±K√(1+0.4/K))/2.
We know x is a positive quantity, so:
x=(-K+K√(1+0.4/K))/2=K(√(1+0.4/K)-1)/2.
0.4/K is a large number, so:
x=K(√(0.4/K)-1)/2=0.00136.
pH=-log(0.00136)=2.866 approx.
More to follow in due course (I hope!)...