(a) The substitution gives \(\mathrm{d}x=\sqrt3\sec^2u\,\mathrm{d}u\) and \(3+x^2=3\sec^2u\). The lower limit \(x=1\) gives \(\tan u=1/\sqrt3\), so \(u=\pi/6\). The upper limit \(x=3\) gives \(\tan u=\sqrt3\), so \(u=\pi/3\).

\[I=\int_{\pi/6}^{\pi/3}\frac{3\sqrt3\tan^3u}{3\sec^2u}\sqrt3\sec^2u\,\mathrm{d}u=\int_{\pi/6}^{\pi/3}3\tan^3u\,\mathrm{d}u.\]

(b) Use \(\tan^3u=\tan u(\sec^2u-1)\):

\[\int3\tan^3u\,\mathrm{d}u=\frac32\tan^2u+3\ln(\cos u).\]

\[I=\frac32\left(3-\frac13\right)+3\ln\left(\frac{1/2}{\sqrt3/2}\right)=\boxed{4-\frac32\ln3}.\]