Answer: \(\mathrm{fg}(x)=\frac{x^2+4}{x^2}\); \(\mathrm g^{-1}(x)=\sqrt{x-1}\); \(x=2\).

(a) Here \(\mathrm{fg}(x)\) means \(\mathrm f(\mathrm g(x))\). Since \(\mathrm g(x)=1+x^2\),

Therefore

\(\mathrm{fg}(x)=\frac{x^2+4}{x^2}.\)

(b) Let

\(y=1+x^2.\)

Then

\(x^2=y-1.\)

Since the domain has \(x\gt 1\), use the positive square root:

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

Hence

\(\mathrm g^{-1}(x)=\sqrt{x-1}.\)

(c) Solve

\(\frac{x+3}{x-1}=1+x^2.\)

Multiplying by \(x-1\),

\(x+3=(x-1)(x^2+1).\)

Expanding,

\(x+3=x^3-x^2+x-1.\)

Cancel \(x\) from both sides and rearrange:

\(x^3-x^2-4=0.\)

Since \(x=2\) is a root, factorise:

\(x^3-x^2-4=(x-2)(x^2+x+2).\)

The quadratic \(x^2+x+2\) has discriminant

\(1-8=-7,\)

so it has no real roots. Therefore the only solution is

\(x=2.\)