Đáp án:

$\begin{array}{l}
a)\left( {{x^2} + c.x + 2} \right)\left( {a.x + b} \right)\\
 = a.{x^3} + b{x^2} + ac{x^2} + bcx + 2ax + 2b\\
 = a{x^3} + \left( {b + ac} \right){x^2} + \left( {bc + 2a} \right)x + 2b\\
 = {x^3} + {x^2} - 2\\
 \Rightarrow \left\{ \begin{array}{l}
a = 1\\
b + ac = 1\\
bc + 2a = 0\\
2b =  - 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 1\\
b =  - 1\\
 - 1 + 1.c = 1\\
 - 1.c + 2.1 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 1\\
b =  - 1\\
c = 2
\end{array} \right.\\
b)\left( {a{y^2} + by + c} \right)\left( {y + 3} \right)\\
 = a{y^3} + 3a{y^2} + b{y^2} + 3by + cy + 3c\\
 = a{y^3} + \left( {3a + b} \right){y^2} + \left( {3b + c} \right)y + 3c\\
 = {y^3} + 2{y^2} - 3y\\
 \Rightarrow \left\{ \begin{array}{l}
a = 1\\
3a + b = 2\\
3b + c =  - 3\\
3c = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 1\\
3 + b = 2\\
3b + c =  - 3\\
c = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
a = 1\\
b =  - 1\\
c = 0
\end{array} \right.\\
c)\left( {{z^2} - z + 1} \right)\left( {a{z^2} + bz + c} \right)\\
 = a{z^4} + b{z^3} + c{z^2} - a{z^3} - b{z^2} - cz\\
 + a{z^2} + bz + c\\
 = a{z^4} + \left( {b - a} \right){z^3} + \left( {c - b + a} \right){z^2}\\
 + \left( { - c + b} \right)z + c\\
 = 2{z^4} - {z^3} + 2{z^2} + 1\\
 \Rightarrow \left\{ \begin{array}{l}
a = 2\\
b - a =  - 1\\
c - b + a = 2\\
 - c + b = 0\\
c = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 2\\
b = 1\\
c - 1 + 2 = 2\\
b = c\\
c = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
a = 2\\
b = c = 1
\end{array} \right.
\end{array}$

Câu 6 thiếu đề bài.