(a) The decrease in kinetic energy (KE) is calculated using the formula:

\(\text{Decrease in KE} = \frac{1}{2} \times 4 \times (12^2 - 8^2)\)

\(= \frac{1}{2} \times 4 \times (144 - 64)\)

\(= \frac{1}{2} \times 4 \times 80\)

\(= 160 \text{ J}\)

(b) The potential energy (PE) gained is:

\(\text{PE gained} = 4g \times 10 \sin 30^{\circ}\)

\(= 4 \times 9.8 \times 10 \times 0.5\)

\(= 200 \text{ J}\)

The total work done is the difference between the potential energy gained and the decrease in kinetic energy:

\(\text{Total work done} = 200 - 160\)

\(= 40 \text{ J}\)

(c) When the force is removed, the block decelerates due to gravity. The acceleration \(a\) is given by:

\(-4g \sin 30^{\circ} = 4a\)

\(a = -5 \text{ m/s}^2\)

Using the equation of motion:

\(-10 = 8t - \frac{1}{2} \times 5t^2\)

Solving for \(t\), we find:

\(t = 4.16 \text{ s}\)