Water boils at 373.15 K at 101.325 kPa, and its enthalpy of vaporisation is 40.7 kJ/mol…

Step 1. The cooker. We want $T_2$ at $P_2 = 150\ \text{kPa}$, so $\dfrac{1}{T_2} = \dfrac{1}{373.15} - \dfrac{8.314 \times \ln(150/101.325)}{40\,700} = 0.0025997\ \text{K}^{-1}$. Then $T_2 = 384.7\ \text{K} = 111.5\ ^\circ\text{C}$.

Step 2. The vapour pressure at $60\ ^\circ\text{C}$. $T_2 = 333.15\ \text{K}$. $\ln\dfrac{P_2}{101.325} = -\dfrac{40\,700}{8.314}\left(\dfrac{1}{333.15} - \dfrac{1}{373.15}\right) = -1.5751$. So $P_2 = 21.0\ \text{kPa}$.

Step 3. The camp. $\dfrac{1}{T_2} = \dfrac{1}{373.15} - \dfrac{8.314 \times \ln(60/101.325)}{40\,700} = 0.0027870\ \text{K}^{-1}$, so $T_2 = 358.8\ \text{K} = 85.7\ ^\circ\text{C}$.