A drone starts from rest and accelerates vertically upward at 3 m...

Velocity of the drone at $t = 4 \text{ s}$ is given by $v = u + at$.

$v = 0 + 3 \times 4 = 12 \text{ m s}^{-1}$

When the bolt detaches, it acquires the velocity of the drone at that instant. Thus, the initial velocity of the bolt is $u_{b} = 12 \text{ m s}^{-1}$ (upward).

After detaching, the bolt falls freely under gravity. Its acceleration is $g = 10 \text{ m s}^{-2}$ (downward).

The time elapsed from detachment to $t = 7 \text{ s}$ is $\Delta t = 7 - 4 = 3 \text{ s}$.

Velocity of the bolt at $t = 7 \text{ s}$ is $v_{b} = u_{b} - g \Delta t$.

$v_{b} = 12 - 10 \times 3 = -18 \text{ m s}^{-1}$

The negative sign indicates downward direction. The speed of the bolt is $|-18| = 18 \text{ m s}^{-1}$.

The magnitude of acceleration of the bolt is $10 \text{ m s}^{-2}$.

Answer: $18 \text{ m s}^{-1}, 10 \text{ m s}^{-2}$