Answer: \(x=2\).

The notation \(\mathrm{fg}(x)\) means \(\mathrm{f}(\mathrm{g}(x))\).

Since

\(\mathrm{g}(x)=\sqrt{x+2},\)

we have

\(\mathrm{fg}(x)=\mathrm{f}(\sqrt{x+2}).\)

Using \(\mathrm{f}(x)=\dfrac{3x}{x+4}\),

\(\mathrm{fg}(x)=\dfrac{3\sqrt{x+2}}{\sqrt{x+2}+4}.\)

Now solve \(\mathrm{fg}(x)=1\):

\(\dfrac{3\sqrt{x+2}}{\sqrt{x+2}+4}=1.\)

Multiply by \(\sqrt{x+2}+4\):

\(3\sqrt{x+2}=\sqrt{x+2}+4.\)

So

\(2\sqrt{x+2}=4,\)

and hence

\(\sqrt{x+2}=2.\)

Squaring both sides gives

\(x+2=4,\)

so

\(x=2.\)

This satisfies the domain condition \(x\gt -2\), so the solution is valid.