Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{ The L'Hopital's rule does not apply}}{\text{.}} \cr
& \left( {\text{b}} \right){\text{ The limit does not exist}}{\text{.}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to 2} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} = \frac{{{e^{3{{\left( 2 \right)}^2} - 12\left( 2 \right) + 12}}}}{{{{\left( 2 \right)}^4} - 16}} = \frac{{{e^0}}}{0} = \frac{1}{0} \cr
& {\text{Then, the L'Hopital's rule does not apply}}{\text{. }} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {6x - 12} \right){e^{3{x^2} - 12x + 12}}}}{{4{x^3}}}{\text{ should not be done}}{\text{.}} \cr
& \cr
& \left( {\text{b}} \right) \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} = \frac{1}{0} = \infty \cr
& {\text{Calculating the side limits}} \cr
& \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} = \frac{{{e^{3{{\left( {{2^ + }} \right)}^2} - 12\left( {{2^ + }} \right) + 12}}}}{{{{\left( {{2^ + }} \right)}^4} - 16}} = \frac{{{e^0}}}{{{0^ + }}} = + \infty \cr
& \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} = \frac{{{e^{3{{\left( {{2^ - }} \right)}^2} - 12\left( {{2^ - }} \right) + 12}}}}{{{{\left( {{2^ - }} \right)}^4} - 16}} = \frac{{{e^0}}}{{{0^ - }}} = - \infty \cr
& \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}} \ne \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{e^{3{x^2} - 12x + 12}}}}{{{x^4} - 16}},{\text{ Then, the limit does not exist}}{\text{.}} \cr} $$