Answer: \(a=-\frac52\), range \(\mathbb{R}\), and \(x=\pm\sqrt{\frac{\mathrm{e}^4-7}{2}}\).

For \(\ln(2x+5)\) to be defined, \(2x+5\gt0\), so \(x\gt-\frac52\). Hence the least possible value of \(a\) is \(-\frac52\).

With this domain, \(2x+5\) can take every positive value, so \(\ln(2x+5)\) can take every real value. Thus the range is \(\mathbb{R}\).

Now

\(\mathrm{fg}(x)=\mathrm{f}(\mathrm{g}(x))=\ln(2(x^2+1)+5)=\ln(2x^2+7).\)

Set this equal to \(4\):

\(\ln(2x^2+7)=4.\)

Therefore \(2x^2+7=\mathrm{e}^4\), so

\(x^2=\frac{\mathrm{e}^4-7}{2}.\)

Hence

\(x=\pm\sqrt{\frac{\mathrm{e}^4-7}{2}}.\)