The standard trick for dealing with $1^\infty$ forms is to take logs; it’s very useful if you don’t see anything slicker. Let
$L=\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\;;$
then
$\begin{align*} \ln L&=\ln\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\\ &=\lim_{n\to\infty}\ln{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\\ &=\lim_{n\to\infty}\left(\frac{n^2+2}{2n+1}\right)\ln\left(\frac{3n^2+2n+1}{3n^2-5}\right)\;, \end{align*}$
where the second step uses the continuity of the log function. This is an $\infty\cdot 0$ form, which you can easily convert to a $\frac00$ form:
$\lim_{n\to\infty}\frac{\ln\left(\frac{3n^2+2n+1}{3n^2-5}\right)}{\frac{2n+1}{n^2+2}}\;.$
Once you know $\ln L$, recovering $L$ is trivial; just remember to do it!