Using this image solve all the required questions please

Mass of water ($m_w$) = 150 g = 0.15 kg
Specific heat capacity of water ($c_w$) = 4200 J kg$^{-1}$ K$^{-1}$
Initial temperature of water ($\theta_{wi}$) = 25 °C
Final temperature of water ($\theta_{wf}$) = 30 °C
Change in temperature of water ($\Delta \theta_w$) = $\theta_{wf} - \theta_{wi}$ = 30 °C - 25 °C = 5 °C

The heat gained by the water is $\boxed{\text{3150 J}}$.

Now, we can answer part (a).
Assuming that all the heat lost by the pendulum bob is gained by the water:

The heat lost by the pendulum bob is $\boxed{\text{3150 J}}$.

Mass of pendulum bob ($m_b$) = 100 g = 0.1 kg
Initial temperature of pendulum bob ($\theta_{bi}$) = 100 °C
Final temperature of pendulum bob ($\theta_{bf}$) = 30 °C
Change in temperature of pendulum bob ($\Delta \theta_b$) = $\theta_{bi} - \theta_{bf}$ = 100 °C - 30 °C = 70 °C
Heat lost by pendulum bob ($Q_b$) = 3150 J (from part a)

The specific heat capacity of the material of the pendulum bob is $\boxed{\text{450 J kg}^{-1} \text{ K}^{-1}}$.