Calculate the molar mass of a gas given its volume at STP and mass, and determine the new volume of a gas after a temperature increase at constant pressure.

At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 L.
Given volume $V = 0.280 \text{ L}$.

$$n = \frac{V}{\text{Molar volume at STP}} = \frac{0.280 \text{ L}}{22.4 \text{ L/mol}}$$ $$n = 0.0125 \text{ mol}$$

Molar mass ($M$) is the mass ($m$) divided by the number of moles ($n$).
Given mass $m = 0.400 \text{ g}$.

$$M = \frac{m}{n} = \frac{0.400 \text{ g}}{0.0125 \text{ mol}}$$ $$M = 32.0 \text{ g/mol}$$

The molar mass of the gas is $\boxed{\text{32.0 g/mol}}$.

2. What is the new volume?

Given initial temperature $T_1 = 25 \,^\circ\text{C}$.

$$T_1 = 25 + 273.15 = 298.15 \text{ K}$$

The temperature increases by $5 \,^\circ\text{C}$.

$$T_2 = (25 + 5) \,^\circ\text{C} = 30 \,^\circ\text{C}$$ $$T_2 = 30 + 273.15 = 303.15 \text{ K}$$

Since pressure is constant, we use Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given initial volume $V_1 = 0.62 \text{ L}$.

$$V_2 = V_1 \left(\frac{T_2}{T_1}\right) = 0.62 \text{ L} \left(\frac{303.15 \text{ K}}{298.15 \text{ K}}\right)$$ $$V_2 = 0.62 \text{ L} \times 1.01677$$ $$V_2 \approx 0.63 \text{ L}$$

The new volume is $\boxed{\text{0.63 L}}$.

3. Calculate the approximate molar mass of a gas.

Given temperature $T = 30 \,^\circ\text{C}$.

$$T = 30 + 273.15 = 303.15 \text{ K}$$

Given pressure $P = 780 \text{ torr}$.

$$P = 780 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} \approx 1.0263 \text{ atm}$$

The ideal gas law can be written as $PM = \rho RT$, where $M$ is molar mass, $\rho$ is density, and $R$ is the ideal gas constant ($0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$).
Given density $\rho = 3.33 \text{ g/L}$.

$$M = \frac{\rho RT}{P}$$ $$M = \frac{(3.33 \text{ g/L})(0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))(303.15 \text{ K})}{1.0263 \text{ atm}}$$ $$M = \frac{82.91 \text{ g}\cdot\text{atm/mol}}{1.0263 \text{ atm}}$$ $$M \approx 80.8 \text{ g/mol}$$

The approximate molar mass of the gas is $\boxed{\text{80.8 g/mol}}$.