Answer: (i) \(f^{-1}(x)=\left(\dfrac{x}{9}\right)^2+1\), domain \(x\gt 0\). (ii) \(45\sqrt2\). (iii) \(x=\dfrac85\).

(i) Let

\(y=9\sqrt{x-1}.\)

Then

\(\frac{y}{9}=\sqrt{x-1}.\)

Square both sides:

\(\left(\frac{y}{9}\right)^2=x-1.\)

So

\(x=\left(\frac{y}{9}\right)^2+1.\)

Swap \(x\) and \(y\):

\(f^{-1}(x)=\left(\frac{x}{9}\right)^2+1.\)

Since \(x\gt 1\) for \(f\), we have \(\sqrt{x-1}\gt 0\), so the range of \(f\) is \(y\gt 0\). Therefore the domain of \(f^{-1}\) is

\(x\gt 0.\)

(ii) First find \(g(7)\):

\(g(7)=7^2+2=51.\)

Therefore

\(fg(7)=f(51)=9\sqrt{51-1}=9\sqrt{50}=45\sqrt2.\)

(iii) We have

\(gf(x)=g(f(x))=(9\sqrt{x-1})^2+2=81(x-1)+2=81x-79.\)

So the equation becomes

\(81x-79=5x^2+83x-95.\)

Rearranging,

\(5x^2+2x-16=0.\)

Factorise:

\((5x-8)(x+2)=0.\)

Hence

\(x=\frac85 \quad\text{or}\quad x=-2.\)

But the domain is \(x\gt 1\), so only

\(\boxed{x=\frac85}\)

is valid.