1. $y=\frac{6 x^{2}+4 x+3}{3 x^{2}+2 x+1}$ The maximum value is $\qquad$
2.1. $\frac{7}{2}$.
Notice that $y=2+\frac{1}{3 x^{2}+2 x+1}$.
And $y=3 x^{2}+2 x+1$ has a minimum value of $\frac{12-4}{12}=\frac{2}{3}$. Therefore, the maximum value sought is $\frac{7}{2}$.