Statement-1 : It is easier to pull a heavy object than to push it on a level ground and Statement-2 : The magnitude of frictional force depends on the nature of the two surfaces in contact.

Step 1: Analyze the forces during pushing.

When pushing an object of mass $m$ with a force $F$ at an angle $\theta$ below the horizontal, the vertical component of the force acts downwards, increasing the normal reaction force.

$$N_{\text{push}} = mg + F\sin\theta$$

Step 2: Analyze the forces during pulling.

When pulling the object with a force $F$ at an angle $\theta$ above the horizontal, the vertical component of the force acts upwards, decreasing the normal reaction force.

$$N_{\text{pull}} = mg - F\sin\theta$$

Step 3: Compare the frictional forces.

The frictional force is directly proportional to the normal reaction force.

$$f = \mu N$$

Since $N_{\text{push}} > N_{\text{pull}}$, the frictional force experienced during pushing is greater than that during pulling.

$$f_{\text{push}} > f_{\text{pull}}$$

Step 4: Determine the minimum force required to move the object.

The minimum force required to overcome friction and move the object in each case is given by the following expressions:

$$F_{\text{push}} = \frac{\mu mg}{\cos\theta - \mu\sin\theta}$$ $$F_{\text{pull}} = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$$

Since $F_{\text{pull}} < F_{\text{push}}$, pulling requires less force than pushing, which makes Statement-1 true.

Step 5: Evaluate Statement-2 and its relationship to Statement-1.

The magnitude of the frictional force depends on the coefficient of friction $\mu$, which is determined by the nature of the two surfaces in contact.

$$f = \mu N$$

This makes Statement-2 true. However, the ease of pulling over pushing is due to the change in the normal force $N$, while the nature of the contact surfaces remains identical in both cases. Thus, Statement-2 is not the correct explanation for Statement-1.

Answer: Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.