Calculate the probabilities for selecting bats with specific weights from a normally distributed population.

For $X_1 = 1 \text{ gram}$:

$$Z_1 = \frac{X_1 - \mu}{\sigma} = \frac{1 - 1.25}{0.25} = \frac{-0.25}{0.25} = -1.00$$

For $X_2 = 2 \text{ grams}$:

$$Z_2 = \frac{X_2 - \mu}{\sigma} = \frac{2 - 1.25}{0.25} = \frac{0.75}{0.25} = 3.00$$

We need to find $P(1 < X < 2)$, which is equivalent to $P(-1.00 < Z < 3.00)$.
Using a standard normal distribution table (Z-table):
$P(Z < 3.00) = 0.9987$
$P(Z < -1.00) = 0.1587$

$$P(-1.00 < Z < 3.00) = P(Z < 3.00) - P(Z < -1.00)$$ $$P(-1.00 < Z < 3.00) = 0.9987 - 0.1587$$ $$P(-1.00 < Z < 3.00) = 0.8400$$

The probability is:

$$\boxed{0.8400}$$

For $X = 2.2 \text{ grams}$:

$$Z = \frac{X - \mu}{\sigma} = \frac{2.2 - 1.25}{0.25} = \frac{0.95}{0.25} = 3.80$$

We need to find $P(X > 2.2)$, which is equivalent to $P(Z > 3.80)$.
Using a standard normal distribution table (Z-table):
$P(Z > 3.80) = 1 - P(Z < 3.80)$
$P(Z < 3.80) \approx 0.9999$ (since 3.80 is a very high Z-score)

$$P(Z > 3.80) = 1 - 0.9999$$ $$P(Z > 3.80) = 0.0001$$

The probability is:

$$\boxed{0.0001}$$