Step 1: Given data

1. Height at which bird is sitting, $h=5m$
2. Time taken is $7:00a.m.$ to $7:10a.m.$, $\therefore t=10minutes=10\times 60=600s$
3. Distance traveled in one trip, $d=2\times h=2\times 5=10m$
4. Total number of trips taken, $n=5$

Step 2: Find the average speed of the bird.

Distance:

1. It is the length of the complete path traveled by a body.
2. It is a scalar quantity as it depends upon the magnitude and not the direction.
3. It can have positive values.

Displacement:

1. It is the shortest path between the initial and final positions.
2. It is a vector quantity as it has a direction and magnitude.
3. It can be positive, negative, and even zero.
4. The formula for average speed is $\mathit{A}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{s}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\mathbf{=}\frac{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$
5. According to the given question, $Averagespeed=\frac{n\times d}{t}$

$$\begin{gathered} =\frac{5\times 10}{600} \\ =0.5m{s}^{-1} \end{gathered}$$

Step 3: Find the average velocity of the bird.

1. The formula for average velocity is $\mathit{A}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{splacement}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$
2. In the given question, the bird is coming down from the tree and covering $5m$ and again flying back to the same height on the tree to build the nest and covers next $5m$ in one trip.
3. After $5$ same trips, the final position of the bird is the same as the initial position .i.e, $5m$ above the ground on the tree.
4. Thus, the total displacement of the bird after $5$ trips is zero.
5. Therefore, the average velocity of the bird is zero.

Thus:

(a) The average speed of the bird is $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.

(b) The average velocity of the bird is $\mathbf{0}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.