How can I determine the pH of solution?
Concerning the first part of your problem: In the general case, which you requested, we start with the mass action law:

$$
K_\text{b} = \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]} = \frac{[\ce{OH-}]^2}{[\ce{B}]}
$$

That’s essentially what you just reformulated to obtain your expression for $\mathrm{pH}$ above. At this point, you made an error in the sign of the second term on the right-hand side – it should be added, not subtracted, i.e. $\mathrm{pH} = 14 + \frac{1}{2} \lg \left( [\ce{B}]K_\text{b} \right)$.

Now comes the important thing: The mass action law uses the concentrations of our components in the equilibrium! What is given as 0.1 M is, however, the starting concentration $[\ce{B}]_0$ before equilibrium, which will of course decrease; $[\ce{HB+}]$ and $[\ce{OH-}]$ will increase by the same amount, which we will denote $x$. Thus, we rewrite:

$$
K_\text{b} = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{x^2}{[\ce{B}]_0 - x}
$$

Now, you just solve for $x$, which is the concentration $[\ce{OH-}]$.

One more thing. You will see that $x$ is actually pretty small in this case, so $[\ce{B}]_0 - x \approx [\ce{B}]_0$. By using $K_\text{b} = \frac{x^2}{[\ce{B}]_0}$, we essentially obtain the same result. You can’t always rely on this, however, and if you have stronger acids or bases, this will get important.