Amphoteric amino acid titration
As in aqueous solution:
$$\ce{NH3+CH2COOH_{[b]}&lt;=&gt; NH3+CH2COO-_{[z]} + H+_{[x]}\\NH3+CH2COO-_{[z]} &lt;=&gt; NH2CH2COO-_{[a]} + H+_{[x]}}$$ 
So,
$$10^{-2}=\frac{[z][x]}{[b]}\\10^{-10}=\frac{[a][x]}{[z]}\\\implies 10^{-12}=[x]^2$$
Notice that a=b as the solution is always neutral; also z is a neutral compund
As, 
$$[H^+]=10^{-6}\implies p^H=6$$
$$\implies[b]=100[z][x]=10^{2-2-6}=10^{-6}$$


  You should have done $p^H=-log[H^+]$, your [b] is correct.




For part (b), yes, using Henderson-Hasslebalch equation:
$$p^H=10+\log\left(\frac{1.05\times10^{-4}}{10^{-6}}\right)\approx 10.97$$



In basic solution, the second equilibrium shifts well(as a result the first one too) to the right.As NaOH is a strong acid, we can neglect any other bases.In presence of NaOH,
$$\ce{NH3+CH2COOH_{[b]} + 2NaOH_{[k]}&lt;=&gt; 2NH2CH2COONa_{[a]} + 2H2O_{[y]}}$$
In millimoles:
$$\begin{array}{c|c|c|c|c}
&amp;entity&amp;b&amp;k&amp;a&amp;y\\\hline
&amp;initial&amp;10&amp;.1&amp;0&amp;0\\\hline
&amp;final&amp;9.95&amp;0&amp;.1&amp;.1+110(original)\\\hline
\end{array}$$
Now you can similiarly apply agin the Henderson-Hasslebalch equation:
$$p_H=12+log(.1/9.95)\approx 10.01$$

Note that I am treating the compund as a monoprotic acid of $pK_a =10(-NH_3^+)+2(-COOH)=12$