Answer: \(a=-1\), and \(x=0.916\).

Rewrite the exponential terms in a common form, or introduce a substitution, so that the equation becomes algebraic.

For \(\mathrm{gf}(x)=\mathrm g(\mathrm f(x))\) to exist, the output of \(\mathrm f\) must be at least \(1\), because the domain of \(\mathrm g\) is \(x\geqslant1\).

Since \(x\geqslant0\), the least value of \(2\mathrm e^x+a\) is \(2+a\). Therefore

\(2+a\geq1.\)

The least integer value is

\(a=-1.\)

When \(a=5\),

\(\mathrm{gf}(x)=\sqrt{(2\mathrm e^x+5)-1}=\sqrt{2\mathrm e^x+4}.\)

Solve \(\mathrm{gf}(x)=3\):

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

Squaring gives

\(2\mathrm e^x+4=9.\)

So

\(\mathrm e^x=\frac52.\)

Hence

\(x=\ln\left(\frac52\right)=0.916\ldots.\)

Therefore \(x=0.916\) to 3 decimal places.

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.