Calculate the molarity of NaOH and the concentration of a diluted FeCl3 solution using the provided data.

Molar mass of KHP $= (1 \times 39.098) + (5 \times 1.008) + (8 \times 12.011) + (4 \times 15.999) = 204.222 \text{ g/mol}$

The reaction between KHP and NaOH is 1:1.

Therefore, moles of NaOH = moles of KHP $= 0.0040319 \text{ mol}$.

The exact molarity of the NaOH solution is $\boxed{\text{0.1040 M}}$.

2. Concentration of the diluted $\text{FeCl}_3$ solution

Initial concentration ($M_1$) $= 3.47 \times 10^{-2} \text{ M}$
Initial volume ($V_1$) $= 5.00 \text{ cm}^3$
Final volume ($V_2$) $= 250 \text{ cm}^3$

The concentration of the diluted solution is $\boxed{\text{6.94} \times 10^{-4} \text{ M}}$.

3. Molarity of the $\text{FeCl}_3$ (aq) solution

Molar mass of $\text{FeCl}_3 = (1 \times 55.845) + (3 \times 35.453) = 55.845 + 106.359 = 162.204 \text{ g/mol}$

The molarity of the $\text{FeCl}_3$ (aq) solution is $\boxed{\text{0.0347 M}}$.

4. Concentration of the diluted solution

Initial concentration ($M_1$) $= 0.03477 \text{ M}$ (using the provided answer from question 3 for consistency)
Initial volume ($V_1$) $= 5.00 \text{ cm}^3$
Final volume ($V_2$) $= 250 \text{ cm}^3$

The concentration of the diluted solution is $\boxed{\text{0.0006954 M}}$.