Answer
The first three terms starting from $n=0$:
(a) $\left( {{a_0}^2,{a_1}^2,{a_2}^2} \right) = \left( {9,4,\dfrac{1}{4}} \right)$
(b) $\left( {{b_0},{b_1},{b_2}} \right) = \left( {0,\dfrac{1}{{24}},\dfrac{1}{{60}}} \right)$
(c) $\left( {{a_0}{b_0},{a_1}{b_1},{a_2}{b_2}} \right) = \left( {0, - \dfrac{1}{{12}}, - \dfrac{1}{{120}}} \right)$
(d) $\left( {2{a_1} - 3{a_0},2{a_2} - 3{a_1},2{a_3} - 3{a_2}} \right) = \left( {5,5,\dfrac{3}{2}} \right)$
Work Step by Step
(a) We have ${a_n} = \dfrac{{n - 3}}{{n!}}$. So, ${a_n}^2 = {\left( {\dfrac{{n - 3}}{{n!}}} \right)^2}$.
The first three terms of ${a_n}^2$ starting from $n=0$ is
$\left( {{a_0}^2,{a_1}^2,{a_2}^2} \right) = \left( {9,4,\dfrac{1}{4}} \right)$
(b) We have ${b_n} = {a_{n + 3}} = \dfrac{n}{{\left( {n + 3} \right)!}}$. So,
The first three terms of ${b_n}$ starting from $n=0$ is
$\left( {{b_0},{b_1},{b_2}} \right) = \left( {0,\dfrac{1}{{24}},\dfrac{1}{{60}}} \right)$
(c)
${a_n}{b_n} = \left( {\dfrac{{n - 3}}{{n!}}} \right)\left( {\dfrac{n}{{\left( {n + 3} \right)!}}} \right) = \dfrac{{n\left( {n - 3} \right)}}{{n!\left( {n + 3} \right)!}}$
The first three terms of ${a_n}{b_n}$ starting from $n=0$ is
$\left( {{a_0}{b_0},{a_1}{b_1},{a_2}{b_2}} \right) = \left( {0, - \dfrac{1}{{12}}, - \dfrac{1}{{120}}} \right)$
(d)
$2{a_{n + 1}} - 3{a_n} = \dfrac{{2\left( {n - 2} \right)}}{{\left( {n + 1} \right)!}} - \dfrac{{3\left( {n - 3} \right)}}{{n!}}$
The first three terms of $2{a_{n + 1}} - 3{a_n}$ starting from $n=0$ is
$\left( {2{a_1} - 3{a_0},2{a_2} - 3{a_1},2{a_3} - 3{a_2}} \right) = \left( {5,5,\dfrac{3}{2}} \right)$
