Define an acid and base based on the following theories:

The concentration of dissociated ions is given by $C \times \alpha$.
$$
[\text{CH}_3\text{COO}^-] = [\text{H}^+] = C \times \alpha = 0.1 \text{ M} \times 0.03 = 0.003 \text{ M}
$$
The concentration of dissociated ions is $\boxed{\text{0.003 M}}$.

The concentration of undissociated acid at equilibrium is $C(1-\alpha)$.
$$
[\text{CH}_3\text{COOH}] = C(1-\alpha) = 0.1 \text{ M} \times (1 - 0.03) = 0.1 \text{ M} \times 0.97 = 0.097 \text{ M}
$$
The concentration of undissociated acid is $\boxed{\text{0.097 M}}$.

$$
\text{pOH} = -\log[\text{OH}^-]
$$
$$
\text{pOH} = -\log(3.0 \times 10^{-3})
$$
$$
\text{pOH} \approx 2.52
$$

$$
\text{pH} = 14 - \text{pOH}
$$
$$
\text{pH} = 14 - 2.52
$$
$$
\text{pH} = 11.48
$$
The pH of the solution is $\boxed{\text{11.48}}$.

$$
\text{pH} = \text{pKa} + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
$$

$$
\text{pH} = 4.74 + \log\left(\frac{0.1 \text{ M}}{0.2 \text{ M}}\right)
$$
$$
\text{pH} = 4.74 + \log(0.5)
$$
$$
\text{pH} = 4.74 + (-0.301)
$$
$$
\text{pH} = 4.439
$$
The pH of the buffer solution is $\boxed{\text{4.44}}$.