Answer
$$\left. {\text{a}} \right) - 1,{\text{ }}\left. {\text{b}} \right) - 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{\sin \left( {{x^2} + 3x + 2} \right)}}{{x + 2}} \cr
& \left. {\text{a}} \right){\text{Completing the table}} \cr
& f\left( { - 2.1} \right) = \frac{{\sin \left( {{{\left( { - 2.1} \right)}^2} + 3\left( { - 2.1} \right) + 2} \right)}}{{\left( { - 2.1} \right) + 2}} = - 1.097 \cr
& f\left( { - 2.01} \right) = \frac{{\sin \left( {{{\left( { - 2.01} \right)}^2} + 3\left( { - 2.01} \right) + 2} \right)}}{{\left( { - 2.01} \right) + 2}} = - 1.009 \cr
& f\left( { - 2.001} \right) = \frac{{\sin \left( {{{\left( { - 2.001} \right)}^2} + 3\left( { - 2.001} \right) + 2} \right)}}{{\left( { - 2.001} \right) + 2}} = - 1.00099 \cr
& f\left( { - 1.999} \right) = \frac{{\sin \left( {{{\left( { - 1.999} \right)}^2} + 3\left( { - 1.999} \right) + 2} \right)}}{{\left( { - 1.999} \right) + 2}} = - 0.9989 \cr
& f\left( { - 1.99} \right) = \frac{{\sin \left( {{{\left( { - 1.99} \right)}^2} + 3\left( { - 1.99} \right) + 2} \right)}}{{\left( { - 1.99} \right) + 2}} = - 0.9899 \cr
& f\left( { - 1.9} \right) = \frac{{\sin \left( {{{\left( { - 1.9} \right)}^2} + 3\left( { - 1.9} \right) + 2} \right)}}{{\left( { - 1.9} \right) + 2}} = - 0.8987 \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
x&{ - 2.1}&{ - 2.01}&{ - 2.001}&{ - 1.999}&{ - 1.99}&{ - 1.9} \\
{f\left( x \right)}&{ - 1.097}&{ - 1.009}&{ - 1.00099}&{ - 0.9989}&{ - 0.9899}&{ - 0.8987}
\end{array}}\]
$$\eqalign{
& {\text{Then we can assume that}} \cr
& \mathop {\lim }\limits_{x \to - 2} \frac{{\sin \left( {{x^2} + 3x + 2} \right)}}{{x + 2}} = - 1 \cr
& \cr
& \left. {\text{b}} \right){\text{Calculating the exact value of the limit}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {{x^2} + 3x + 2} \right)}}{{x + 2}} = \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x + 2} \right)\left( {x + 1} \right)}}{{x + 2}} \cr
& {\text{Multiplying by }}\frac{{x + 1}}{{x + 1}} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x + 1} \right)\sin \left( {x + 2} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} \cr
& {\text{Using the property of product of limits}} \cr
& = \left[ {\mathop {\lim }\limits_{x \to - 2} \left( {x + 1} \right)} \right]\left[ {\mathop {\lim }\limits_{x \to - 2} \frac{{\sin \left( {x + 2} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}} \right] \cr
& {\text{Calculate}} \cr
& = \left( { - 2 + 1} \right)\left( 1 \right) \cr
& = \left( { - 1} \right)\left( 1 \right) \cr
& = - 1 \cr
& \cr} $$
