The resultant velocity of the swimmer is the vector sum of the river's velocity and the swimmer's velocity relative to the water.

Given:
- River velocity = 6 km/h
- Swimmer's velocity relative to water = 9 km/h

Using the Pythagorean theorem:

\[
v_{\text{resultant}} = \sqrt{(9^2 + 6^2)} = \sqrt{81 + 36} = \sqrt{117}
\]

Thus, the resultant velocity of the swimmer is:

\[
{\sqrt{117} \text{km/h}}
\]

To find the velocity of the monkey as observed by a man standing on the ground, we need to add the velocity of the monkey relative to the train to the velocity of the train.

 Given:
- Velocity of the train (north direction) = 54 km/h
- Velocity of the monkey relative to the train (opposite to train's motion) = 18 km/h

Monkey's velocity relative to the ground:
Since the monkey is running against the train’s motion, the monkey's velocity relative to the ground will be:

\[
\vec{v}_{\text{monkey}} = \vec{v}_{\text{train}} - \vec{v}_{\text{monkey relative to train}}
\]

\[
v_{\text{monkey}} = 54 \, \text{km/h} - 18 \, \text{km/h} = 36 \, \text{km/h}
\]

Thus, the velocity of the monkey as observed by a man on the ground is:

\[
{36 \, \text{km/h} \text{ north} or 10 m/s \text{ north}}
\]

The relative velocity of the boat with respect to the water is given by subtracting the velocity of the water from the velocity of the boat.

\[
\vec{v}_{\text{bw}} = \vec{v}_{\text{boat}} - \vec{v}_{\text{water}}
\]

Given:
\[
\vec{v}_{\text{boat}} = 3\hat{i} + 4\hat{j}, \quad \vec{v}_{\text{water}} = -3\hat{i} - 4\hat{j}
\]

Now, subtract:

\[
\vec{v}_{\text{bw}} = (3\hat{i} + 4\hat{j}) - (-3\hat{i} - 4\hat{j})
\]
\[
\vec{v}_{\text{bw}} = 3\hat{i} + 4\hat{j} + 3\hat{i} + 4\hat{j}
\]
\[
\vec{v}_{\text{bw}} = 6\hat{i} + 8\hat{j}
\]

Thus, the relative velocity of the boat with respect to the water is:

\[ {6\hat{i} + 8\hat{j}} \]