Calculate the pK_a of lactic acid, given that when the concentration of lactic acid is 0.010 M and the concentration of lactate is 0.087 M, the pH is 4.80.

$$
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right)
$$
Here, lactic acid is the weak acid and lactate is its conjugate base.

$$
4.80 = \text{pK}_a + \log\left(\frac{0.087 \text{ M}}{0.010 \text{ M}}\right)
$$

$$
4.80 = \text{pK}_a + \log(8.7)
$$
$$
4.80 = \text{pK}_a + 0.9395
$$

$$
\text{pK}_a = 4.80 - 0.9395
$$
$$
\text{pK}_a = 3.8605
$$
The pK$_a$ of lactic acid is $\boxed{3.86}$.

$$
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right)
$$

$$
\text{pH} = 4.76 + \log\left(\frac{0.20 \text{ M}}{0.10 \text{ M}}\right)
$$

$$
\text{pH} = 4.76 + \log(2.0)
$$
$$
\text{pH} = 4.76 + 0.3010
$$
$$
\text{pH} = 5.0610
$$
The pH of the mixture is $\boxed{5.06}$.

$$
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{acetate}]}{[\text{acetic acid}]}\right)
$$
Given $\text{pH} = 5.30$ and $\text{pK}_a = 4.76$.

$$
5.30 = 4.76 + \log\left(\frac{[\text{acetate}]}{[\text{acetic acid}]}\right)
$$

$$
\log\left(\frac{[\text{acetate}]}{[\text{acetic acid}]}\right) = 5.30 - 4.76
$$
$$
\log\left(\frac{[\text{acetate}]}{[\text{acetic acid}]}\right) = 0.54
$$

$$
\frac{[\text{acetate}]}{[\text{acetic acid}]} = 10^{0.54}
$$
$$
\frac{[\text{acetate}]}{[\text{acetic acid}]} = 3.467
$$
The ratio of the concentrations of acetate to acetic acid is $\boxed{3.47}$.

$$
[\text{weak acid}] = [\text{acetic acid}] = 0.2 \text{ M}
$$
$$
[\text{conjugate base}] = [\text{acetate ion}] = 2 \text{ M}
$$
$$
\text{pK}_a = 4.8
$$

$$
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right)
$$

$$
\text{pH} = 4.8 + \log\left(\frac{2 \text{ M}}{0.2 \text{ M}}\right)
$$

$$
\text{pH} = 4.8 + \log(10)
$$
$$
\text{pH} = 4.8 + 1
$$
$$
\text{pH} = 5.8
$$
The approximate pH of the solution is $\boxed{5.8}$.