Answer: (a) \(\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}\); (b) \(AC=\frac{\sqrt6+\sqrt2}{2}\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use the quadrant and the exact trigonometric ratio to select the correct solutions.

(a) In triangle \(ABC\),

\(\angle A=60^\circ,\quad \angle B=75^\circ,\quad \angle C=45^\circ,\)

and

\(AB=\sqrt2.\)

Using the sine rule,

\(\frac{BC}{\sin60^\circ}=\frac{AB}{\sin45^\circ}.\)

Therefore

\(\frac{BC}{\sin60^\circ}=\frac{\sqrt2}{\sin45^\circ}.\)

Since \(\sin45^\circ=\frac{\sqrt2}{2}\),

\(\frac{\sqrt2}{\sin45^\circ}=2.\)

So

\(BC=2\sin60^\circ=2\cdot\frac{\sqrt3}{2}=\sqrt3.\)

The area of the triangle can also be written using sides \(AB\) and \(BC\):

\(\text{area}=\frac12(AB)(BC)\sin75^\circ.\)

Substitute \(AB=\sqrt2\) and \(BC=\sqrt3\):

\(\text{area}=\frac12\sqrt2\sqrt3\sin75^\circ=\frac{\sqrt6}{2}\sin75^\circ.\)

The given area is \(\frac{3+\sqrt3}{4}\), so

\(\frac{\sqrt6}{2}\sin75^\circ=\frac{3+\sqrt3}{4}.\)

Hence

\(\sin75^\circ=\frac{3+\sqrt3}{2\sqrt6}.\)

Rationalising and simplifying,

\(\sin75^\circ=\frac{(3+\sqrt3)\sqrt6}{12} =\frac{3\sqrt6+\sqrt{18}}{12} =\frac{3\sqrt6+3\sqrt2}{12}.\)

Therefore

\(\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}.\)

(b) Use the sine rule again:

\(\frac{AC}{\sin75^\circ}=\frac{AB}{\sin45^\circ}.\)

As before,

\(\frac{AB}{\sin45^\circ}=2.\)

So

\(AC=2\sin75^\circ.\)

Using part (a),

\(AC=2\cdot\frac{\sqrt6+\sqrt2}{4} =\frac{\sqrt6+\sqrt2}{2}.\)

This completes the solution and gives the required result.