Finding pH of a buffer What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine, $\ce{CH3NH2}$,
with 25.00 mL of 0.10 M methylammonium chloride, $\ce{CH3NH3Cl}$? Assume that the volume of
the solutions are additive and that Kb = 3.70 × 10-4 for methylamine.

1) First, I multiplied 0.025 L with 0.10 M and received 0.0025 moles. I then divided 0.0025 moles with 0.050 L (The total volume of the solution) and received 0.05 M, the molar concentration of the methylamine and its conjugate acid in this mixture.

2) I then created an ICE Chart (my reactant was the methylamine so that I would not need to find the Ka) and I solved for "x" (which would give me the concentration of $\ce{OH}$ ions at equilibrium). First, I solved for "x" but I assumed that is was insignificant. In the end I received 10.57 as the pH, which was the wrong answer. Then when I solved for "x" using the quadratic formula (assuming that "x" was significant), I got 0.0003646 as my "x". 

3) When I took the negative log of 0.0003646, I got a pOH of 3.46. I subtracted this number by 14, and I still received a pH of 10.57

4) I even attempted to use the Henderson Hasselbalch equation and still received 10.57 because the ratio of the base and acid is 1, meaning that the log is 0.

I don't understand what I am doing incorrect. The answer is supposed to be 10.87, can someone please assist me?
Your approach is correct and the result you obtain is right. Checked here. 

As you correctly affirm, since the ratio of the base and acid is $1$, log will be zero and therefore $\mathrm{pH}=\mathrm{p}K_\text{a}$, where $\mathrm{p}K_\text{a} = 10.57$ (derived from the value of $K_\text{b}$ you provide).