Given the following experimental data, determine the reaction orders m and n for the reaction rate = -kf CA^m C_Bⁿ.

The rate law is $rate = -k_f C_A^m C_B^n$.
Using the data from Experiment 1 and Experiment 2:

$$\frac{Rate_2}{Rate_1} = \frac{-k_f C_{A,2}^m C_{B,2}^n}{-k_f C_{A,1}^m C_{B,1}^n}$$

Substitute the values:

$$\frac{-4.90 \times 10^{-4} \text{ mol/L.s}}{-2.45 \times 10^{-4} \text{ mol/L.s}} = \frac{(2 \text{ mol/L})^m (1 \text{ mol/L})^n}{(1 \text{ mol/L})^m (1 \text{ mol/L})^n}$$ $$2 = \left(\frac{2}{1}\right)^m \left(\frac{1}{1}\right)^n$$ $$2 = 2^m \cdot 1^n$$ $$2 = 2^m$$

Therefore, $m = 1$.
The order of reaction with respect to A is $\boxed{1}$.

Using the data from Experiment 2 and Experiment 3:

$$\frac{Rate_3}{Rate_2} = \frac{-k_f C_{A,3}^m C_{B,3}^n}{-k_f C_{A,2}^m C_{B,2}^n}$$

Substitute the values and the value of $m=1$:

$$\frac{-4.90 \times 10^{-4} \text{ mol/L.s}}{-4.90 \times 10^{-4} \text{ mol/L.s}} = \frac{(1 \text{ mol/L})^1 (4 \text{ mol/L})^n}{(2 \text{ mol/L})^1 (1 \text{ mol/L})^n}$$ $$1 = \frac{1 \cdot 4^n}{2 \cdot 1}$$ $$1 = \frac{4^n}{2}$$ $$2 = 4^n$$

Since $4 = 2^2$, we can write:

$$2^1 = (2^2)^n$$ $$2^1 = 2^{2n}$$

Equating the exponents:

$$1 = 2n$$ $$n = \frac{1}{2}$$

The order of reaction with respect to B is $\boxed{\frac{1}{2}}$.