1. Interpret the arc division

Since arc $AC$ subtends a right angle at the centre $O$, we have $\angle AOC=90^circ=\frac{\pi}{2}.$

Given $\frac{\text{arc }AB}{\text{arc }BC}=\frac{1}{5},$ the corresponding central angles are in the same ratio. So if $\angle AOB : \angle BOC = 1:5,$ and their sum is $90^circ$, then $\angle AOB = \frac{1}{6}\cdot 90^circ = 15^circ, \qquad \angle BOC=75^circ.$

2. Choose coordinates/vectors

Let the radius of the circle be $r$. Take $\overrightarrow{OA}=r\hat{i}.$ Since $\angle AOC=90^circ$, take $\overrightarrow{OC}=r\hat{j}.$ Point $B$ lies on the arc between $A$ and $C$, with $\angle AOB=15^circ$, so $\overrightarrow{OB}=r(\cos15^circ\,\hat{i}+\sin15^circ\,\hat{j}).$

Now use $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}.$ Substituting, $r\hat{j}=\alpha(r\hat{i})+\beta r(\cos15^circ\,\hat{i}+\sin15^circ\,\hat{j}).$ Dividing by $r$, $\hat{j}=\alpha\hat{i}+\beta\cos15^circ\,\hat{i}+\beta\sin15^circ\,\hat{j}.$

Equating components: $\alpha+\beta\cos15^circ=0 \qquad ...(1)$ $\beta\sin15^circ=1 \qquad ...(2)$

3. Find $\beta$ and $\alpha$

From (2), $\beta=\frac{1}{\sin15^circ}.$ Also from (1), $\alpha=-\beta\cos15^circ=-\frac{\cos15^circ}{\sin15^circ}=-\cot15^circ.$

Now use exact values: $\sin15^circ=\frac{\sqrt6-\sqrt2}{4}, \qquad \cos15^circ=\frac{\sqrt6+\sqrt2}{4}.$ Thus $\beta=\frac{1}{\sin15^circ}=\frac{4}{\sqrt6-\sqrt2}=\sqrt6+\sqrt2,$ using rationalization.

And $\alpha=-\cot15^circ=-\frac{\cos15^circ}{\sin15^circ}=-\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}=-(2+\sqrt3).$

4. Compute the required expression

We need $\alpha+\sqrt2(\sqrt3-1)\beta.$ Now $\beta=\sqrt6+\sqrt2=\sqrt2(\sqrt3+1).$ So