Answer: \(Q=\left(\dfrac{32}{5},13\right)\). The shaded area is \(\dfrac{128}{15}\).

(i) The line is

\(4y=5x+20 \quad\Rightarrow\quad y=\frac54x+5.\)

At the intersection point \(Q\),

\(5+\sqrt{10x}=\frac54x+5.\)

So

\(\sqrt{10x}=\frac54x.\)

For \(x\gt 0\), divide by \(\sqrt x\):

\(\sqrt{10}=\frac54\sqrt x.\)

Hence

\(\sqrt x=\frac{4\sqrt{10}}{5}.\)

Squaring gives

\(x=\frac{16\cdot 10}{25}=\frac{32}{5}.\)

Then

\(y=5+\sqrt{10\cdot \frac{32}{5}}=5+\sqrt{64}=13.\)

Therefore

\(\boxed{Q=\left(\frac{32}{5},13\right)}.\)

(ii) The shaded area is the area between the curve and the line from \(x=0\) to \(x=\dfrac{32}{5}\).

So

\(\text{Area} = \int_0^{32/5}\left[\left(5+\sqrt{10x}\right)-\left(\frac54x+5\right)\right]\,dx.\)

This simplifies to

\(\text{Area} = \int_0^{32/5}\left(\sqrt{10x}-\frac54x\right)\,dx.\)

Integrate term by term:

\(\int \sqrt{10x}\,dx = \sqrt{10}\int x^{1/2}\,dx = \frac{2\sqrt{10}}{3}x^{3/2},\)

and

\(\int \frac54x\,dx=\frac58x^2.\)

Therefore

\(\text{Area} = \left[\frac{2\sqrt{10}}{3}x^{3/2}-\frac58x^2\right]_0^{32/5}.\)

Now from part (i),

\(\sqrt{\frac{32}{5}}=\frac{4\sqrt{10}}{5}.\)

So

\(\sqrt{10}\left(\frac{32}{5}\right)^{3/2} = \sqrt{10}\cdot \frac{32}{5}\cdot \frac{4\sqrt{10}}{5} = \frac{256}{5}.\)

Hence

\(\frac{2\sqrt{10}}{3}\left(\frac{32}{5}\right)^{3/2} = \frac{2}{3}\cdot \frac{256}{5} = \frac{512}{15}.\)

Also,

\(\frac58\left(\frac{32}{5}\right)^2 = \frac58\cdot \frac{1024}{25} = \frac{128}{5} = \frac{384}{15}.\)

So the shaded area is

\(\frac{512}{15}-\frac{384}{15}=\frac{128}{15}.\)

Therefore

\(\boxed{\text{Area}=\frac{128}{15}}.\)