$\text{a)}$

$\{\begin{array}{l}w\left(0\right)=5\\ w\left(1\right)=-3\end{array}$

$\{\begin{array}{l}-2\cdot 0^3+3\cdot 0^2+a\cdot 0+b=5\\ -2\cdot 1^3+3\cdot 1^2+a\cdot 1+b=-3\end{array}$

$\{\begin{array}{l}b=5\\ -2+3+a+b=-3\end{array}$

$\{\begin{array}{l}b=5\\ a+b=-4\end{array}$  

Podstawiamy b=5 do drugiego równania w układzie.

$\{\begin{array}{l}b=5\\ a+5=-4\end{array}$

$\{\begin{array}{l}b=5\\ a=-9\end{array}$  

$\{\begin{array}{l}a=-9\\ b=5\end{array}$  

---

$\text{b)}$  

$\{\begin{array}{l}w\left(-3\right)=11\\ w\left(2\right)=\frac{8}{3}\end{array}$

$\{\begin{array}{l}a\cdot {\left(-3\right)}^3+{\left(-3\right)}^2+b\cdot \left(-3\right)+2=11\\ a\cdot 2^3+2^2+b\cdot 2+2=\frac{8}{3}\end{array}$

$\{\begin{array}{l}-27a+9-3b+2=11\\ 8a+4+2b+2=\frac{8}{3}\mid \cdot 3\end{array}$

$\{\begin{array}{l}-27a-3b=0\\ 24a+6b+18=8\end{array}$

$\{\begin{array}{l}3b=-27a\mid :3\\ 24a+6b=-10\mid :2\end{array}$

$\{\begin{array}{l}b=-9a\\ 12a+3b=-5\end{array}$

Podstawiamy b=-9a do drugiego równania w układzie.

$\{\begin{array}{l}b=-9a\\ 12a+3\cdot \left(-9a\right)=-5\end{array}$  

$\{\begin{array}{l}b=-9a\\ 12a-27a=-5\end{array}$  

$\{\begin{array}{l}b=-9a\\ -15a=-5\mid :\left(-15\right)\end{array}$  

$\{\begin{array}{l}b=-9a\\ a=\frac{1}{3}\end{array}$

Podstawiamy a=¹/₃ do pierwszego równania w układzie.

$\{\begin{array}{l}b=-9\cdot \frac{1}{3}\\ a=\frac{1}{3}\end{array}$

$\{\begin{array}{l}b=-3\\ a=\frac{1}{3}\end{array}$  

$\{\begin{array}{l}a=\frac{1}{3}\\ b=-3\end{array}$  

---

$\text{c)}$

$\{\begin{array}{l}w\left(2\right)=-w\left(1\right)\\ w\left(-2\right)=16\end{array}$

$\{\begin{array}{l}{2}^4+a\cdot 2^3+b\cdot 2^2-10=-\left(1^4+a\cdot 1^3+b\cdot 1^2-10\right)\\ {\left(-2\right)}^4+a\cdot {\left(-2\right)}^3+b\cdot {\left(-2\right)}^2-10=16\end{array}$

$\{\begin{array}{l}16+8a+4b-10=-\left(1+a+b-10\right)\\ 16-8a+4b-10=16\end{array}$

$\{\begin{array}{l}6+8a+4b=9-a-b\\ -8a+4b=10\end{array}$

$\{\begin{array}{l}9a+5b=3\\ 4b=10+8a\mid :4\end{array}$

$\{\begin{array}{l}9a+5b=3\\ b=\frac{5}{2}+2a\end{array}$

Podstawiamy b=⁵/₂+2a do pierwszego równania w układzie.

$\{\begin{array}{l}9a+5\left(\frac{5}{2}+2a\right)=3\\ b=\frac{5}{2}+2a\end{array}$

$\{\begin{array}{l}9a+\frac{25}{2}+10a=\frac{6}{2}\\ b=\frac{5}{2}+2a\end{array}$

$\{\begin{array}{l}19a=-\frac{19}{2}\mid \cdot \frac{1}{19}\\ b=\frac{5}{2}+2a\end{array}$

$\{\begin{array}{l}a=-\frac{1}{2}\\ b=\frac{5}{2}+2a\end{array}$

Podstawiamy a=-¹/₂ do drugiego równania w układzie.

$\{\begin{array}{l}a=-\frac{1}{2}\\ b=\frac{5}{2}+2\cdot \left(-\frac{1}{2}\right)\end{array}$  

$\{\begin{array}{l}a=-\frac{1}{2}\\ b=\frac{5}{2}-\frac{2}{2}\end{array}$  

$\{\begin{array}{l}a=-\frac{1}{2}\\ b=\frac{3}{2}\end{array}$