The titration of 25.00ml of 0.1000m hypochlorous acid (3.5x10^-8) with 0.1500m NaOH.

I am sure you meant to write M and not m. Remember that M stands for molarity (mols/L of solution) and m stands for molality (mols/kg of solvent). I will go with M. Also, if you are titrating HClO WITH NaOH, your first question should be what is the pH BEFORE any NaOH (base) is added.
HClO + NaOH ==> NaClO +H2O
At the beginning you have added no NaOH; therefore, you have a solution of HClO which is 0.1 M. So the pH is determined by the ionization of HClO.
..............HClO ==> H^+ + ClO^-
I............0.1M..........0............0
C..............-x............x.............x
E...........0.1-x..........x.............x
Ka = 3.5E-8 = (H^+)(ClO^-)/(HClO)
Plug the E line into the Ka expression and solve for (H^+) then convert to pH.

Note: If you really meant to determine the pH of NaOH before mixing with HClO, then pH of the NaOH is the pH of 0.15 M NaOH. The OH^- will be 0.15 M, Convert that to pOH, then to pH.

Part B. When 25 mL of 0.15 M NaOH has been added to 25 mL of 0.1 M HClO, you have this. mols HClO = M x L = 0.1 x 0.025 = 0.0025.
mols NaOH = M x L = 0.15 x 0.025 = 0.00375
So you have enough NaOH to neutralize all of the HClO and have an exces. How much excess NaOH do you have? That's 0.00375-0.0025 = ? = mole OH^-. (OH^-) = mols/L . You have 0.00125 mol OH and L = 25 mL + 25 mL = 50 mL = 0.050 L
Then convert (OH^-) to pOH then to pH.
Post your work if you get stuck.