Let Case 1 be the pulling force and Case 2 be the pushing force. The displacement is $d$.

[Q1] The work done by the applied force $F$ is given by $W_F = \vec{F} \cdot \vec{d} = Fd\cos\theta$. Since $F$, $d$, and $\theta$ are the same for both cases, the work done is equal.

$$W_{F1} = W_{F2} = Fd\cos\theta$$

[Q2] The work done by friction is $W_f = -f_k d$, where $f_k = \mu_k N$. The normal force $N$ depends on the vertical component of the applied force. Case 1 (Pull): Vertical equilibrium gives $N_1 + F\sin\theta = mg$, so $N_1 = mg - F\sin\theta$. Case 2 (Push): Vertical equilibrium gives $N_2 = mg + F\sin\theta$. Since $N_2 > N_1$, the friction force is greater when pushing: $f_{k2} > f_{k1}$. Therefore, the work done by friction is not equal; more negative work is done by friction when pushing: $|W_{f2}| > |W_{f1}|$.

[Q3] According to the work-energy theorem, the change in kinetic energy is equal to the net work done: $\Delta K = W_{net} = W_F + W_f$.

$$W_{net1} = Fd\cos\theta - f_{k1}d = Fd\cos\theta - \mu_k(mg - F\sin\theta)d$$ $$W_{net2} = Fd\cos\theta - f_{k2}d = Fd\cos\theta - \mu_k(mg + F\sin\theta)d$$

Since $W_{net1} > W_{net2}$ and the object starts from rest ($\Delta K = \frac{1}{2}mv_f^2$), the final kinetic energy is greater in Case 1. Thus, the final velocities are not equal: $v_{f1} > v_{f2}$.

[Q4] The average power is $P_{avg} = W_F / t$. The work done by the applied force, $W_F$, is the same in both cases. We need to compare the time taken, $t$. The net horizontal force is $F_{net,x} = F\cos\theta - f_k$. Since $f_{k1} < f_{k2}$, the net force is greater in Case 1: $F_{net,x1} > F_{net,x2}$. This means the acceleration is greater in Case 1: $a_1 > a_2$. For motion with constant acceleration over a distance $d$ from rest, $d = \frac{1}{2}at^2$, so $t = \sqrt{2d/a}$. Since $a_1 > a_2$, the time taken is shorter in Case 1: $t_1 < t_2$. Because the work done $W_F$ is the same and $t_1 < t_2$, the average power is greater in Case 1.

$$P_{avg1} = \frac{W_F}{t_1} > \frac{W_F}{t_2} = P_{avg2}$$