Calculate the pH for 0.001 M HCl.

HCl is a strong monoprotic acid, so it dissociates completely.

$$[\text{H}^+] = 0.001 \text{ M} = 10^{-3} \text{ M}$$ $$\text{pH} = -\log[\text{H}^+] = -\log(10^{-3}) = 3$$

H₂SO₄ is a strong diprotic acid. For dilute solutions, both protons are considered to dissociate completely.

$$\text{H}_2\text{SO}_4 \to 2\text{H}^+ + \text{SO}_4^{2-}$$

So, for 0.001 M H₂SO₄, the concentration of $\text{H}^+$ is approximately $2 \times 0.001 \text{ M} = 0.002 \text{ M} = 2 \times 10^{-3} \text{ M}$.

$$\text{pH} = -\log(2 \times 10^{-3}) = 3 - \log(2) \approx 3 - 0.301 = 2.699$$

Since $3 \neq 2.699$, the pH values are not the same.
Therefore, Statement I is incorrect.

Statement II: The pOH of 0.001 M NaOH(aq) is 3.

NaOH is a strong base, so it dissociates completely.

$$\text{NaOH} \to \text{Na}^+ + \text{OH}^-$$

For 0.001 M NaOH, $[\text{OH}^-] = 0.001 \text{ M} = 10^{-3} \text{ M}$.

$$\text{pOH} = -\log[\text{OH}^-] = -\log(10^{-3}) = 3$$

Therefore, Statement II is correct.

Statement III: The ionic product of water, Kw, is temperature dependent.

The autoionization of water is an endothermic process:

$$\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}$$

According to Le Chatelier's principle, an increase in temperature shifts the equilibrium to the right, increasing the concentrations of both $\text{H}^+$ and $\text{OH}^-$. Since $K_w = [\text{H}^+][\text{OH}^-]$, $K_w$ increases with temperature.
Therefore, Statement III is correct.

Statement IV: The pKb of 0.010 M NH₃ solution is greater than 7. (kb of NH₃ = 1.8 x 10⁻⁵)

The pKb is a constant for a given base and does not depend on its concentration.

$$\text{pKb} = -\log(K_b)$$ $$\text{pKb} = -\log(1.8 \times 10^{-5})$$ $$\text{pKb} = 5 - \log(1.8)$$ $$\text{pKb} \approx 5 - 0.255 = 4.745$$

Since 4.745 is not greater than 7, the statement is incorrect.
Therefore, Statement IV is incorrect.

Conclusion:
Statements II and III are correct.

The final answer is $\text{II and III}$.