Given that mn = 2m − n, where m and n are real numbers, find the value of (i) 5(45), (ii) n if n(3*n) = 6. (b) In the diagram, ABD is a straight line. O is the centre of the circle ABC, <OBC = 20°, <ADC = 32° and |CB| = |BD|, find <ACB.

Using the definition $m * n = 2m - n$, with $m=4$ and $n=5$:

Using the definition $m * n = 2m - n$, with $m=5$ and $n=3$:

The value is $\boxed{\text{7}}$.

Using the definition $m * n = 2m - n$, with $m=3$ and the second operand as $n$:

$n * (6 - n) = 6$.
Now, apply the definition $m * n = 2m - n$ again, where the first operand is $n$ and the second operand is $(6 - n)$:

The value of $n$ is $\boxed{\text{4}}$.

10. (b) In the diagram, $ABD$ is a straight line. $O$ is the centre of the circle $ABC$, $\angle OBC = 20^\circ$, $\angle ADC = 32^\circ$ and $|CB| = |BD|$, find $\angle ACB$.

Since $O$ is the center of the circle, $OB = OC$ (radii). Therefore, $\triangle OBC$ is an isosceles triangle.
The base angles are equal: $\angle OCB = \angle OBC = 20^\circ$.
The angle at the center $\angle BOC = 180^\circ - (\angle OBC + \angle OCB) = 180^\circ - (20^\circ + 20^\circ) = 180^\circ - 40^\circ = 140^\circ$.

The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle. For arc $BC$, $\angle BAC$ is the inscribed angle and $\angle BOC$ is the central angle.

Given that $|CB| = |BD|$, $\triangle CBD$ is an isosceles triangle.
The base angles are equal: $\angle BCD = \angle BDC$.
We are given $\angle ADC = 32^\circ$, which is the same as $\angle BDC$.
So, $\angle BCD = 32^\circ$.
The angle $\angle CBD = 180^\circ - (\angle BCD + \angle BDC) = 180^\circ - (32^\circ + 32^\circ) = 180^\circ - 64^\circ = 116^\circ$.

Since $ABD$ is a straight line, $\angle ABC$ and $\angle CBD$ are supplementary angles.

In $\triangle ABC$, the sum of angles is $180^\circ$.

The value of $\angle ACB$ is $\boxed{\text{46}^\circ}$.