A circle with center O has a radius of 5 cm. A chord PQ has a length of 6 cm. Calculate (i) the distance of the chord from the center O, and (ii) the angle POQ.

Let $M$ be the midpoint of the chord $PQ$. The line segment $OM$ represents the perpendicular distance from the center $O$ to the chord $PQ$.
In a circle, the line from the center to the midpoint of a chord is perpendicular to the chord. Therefore, $\triangle OMP$ is a right-angled triangle with the right angle at $M$.

The hypotenuse $OP$ is the radius of the circle, so $OP = 5$ cm.
The side $PM$ is half the length of the chord $PQ$.

The side $OM$ is the distance we need to find.

In $\triangle OMP$:

Substitute the known values:

The distance of the chord from the center $O$ is $\boxed{4 \text{ cm}}$.

The angle $\angle POQ$ is the angle subtended by the chord $PQ$ at the center $O$.
Since $OM$ bisects $\angle POQ$, we have $\angle POQ = 2 \times \angle POM$.
Let $\angle POM = \alpha$.

In the right-angled triangle $\triangle OMP$:
We know $OP = 5$ cm (hypotenuse), $PM = 3$ cm (opposite to $\alpha$), and $OM = 4$ cm (adjacent to $\alpha$).
We can use the sine function:

Using a calculator:

Rounding to one decimal place:

The angle $POQ$ is $\boxed{73.7^\circ}$.