calculating pH - is my answer correct?
The general method to solve problems of this type can be found in the answer here, How to set up equation for buffer reaction?. In your case you need to modify this equation. To follow the text below you need to look at the main result in the linked answer (rather than repeat all this here). The derivation is straightforward although the equation is a bit complicated.

The base constant $K_\mathrm{B}$ is given but as $K_\mathrm{w}=K_\mathrm{A}K_\mathrm{B}$ we can start with  main eqn in linked page and convert to $k_\mathrm{B}$ later on. In this case as ammonia is a base (with concentration $c_b= 0.1$ M) we may suppose that any acid concentration $c_\mathrm{a} \approx 0$ then, 

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} + \mathrm{[H^+]} - \mathrm{[OH^-])} }{(- \mathrm{[H^+]} + \mathrm{[OH^-]} )}  $$ 

Additionally we can reasonably assume that $\mathrm{[OH^-] \gg [H^+]}$ and so 

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} - \mathrm{[OH^-])} }{ \mathrm{[OH^-]} }  $$

This can be solved by substituting for $K_\mathrm{A}=K_\mathrm{w}/K_\mathrm{B}$  and for $\ce{[H^+]}$ as $K_\mathrm{w}=\ce{[H^+][OH^-]}$ and solve the quadratic equation.  But as $c_\mathrm{b} \gg \ce{[OH^-]}$ we can simplify by letting $c_\mathrm{b} - \ce{[OH^-]} \to c_\mathrm{b}$ then the equation becomes  

$$K_\mathrm{A}= \mathrm{[H^+]} \frac{c_\mathrm{b}  }{ \mathrm{[OH^-]} }  $$

We can now get 

$$\frac{K_\mathrm{w}}{K_\mathrm{B}}= \frac{K_\mathrm{w}c_\mathrm{b}} { \mathrm{[OH^-]^2}}   $$

from which 

$$\ce{[OH]^- } =\sqrt{K_\mathrm{b}c_\mathrm{b}}$$

and so you can obtain the pH.

When you dilute, the concentration of base $c_\mathrm{b}$ is halved.