$a)$  

$\text{algebraicznie}$  

$2-x^2=0\iff x\in \left\{-\sqrt{2},\sqrt{2}\right\}$  

$\text{Rozpatrujemy dwa przypadki:}$  

${}^{\text{(1)}}x\in \left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},+\infty \right)\vee {}^{\text{(2)}}x\in ⟨-\sqrt{2},\sqrt{2}⟩$  

$\left(1\right)x\in \left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},+\infty \right)\Rightarrow x^2>2\Rightarrow 2-x^2<0$  

$-\left(2-x^2\right)<x>\frac{<}{x}>\frac{<}{s}pan>\frac{<}{p}><p><spanclass=\text{AM}>‘-2+x^2<x>\frac{<}{x}>\frac{<}{s}pan>\frac{<}{p}><p><spanclass=\text{AM}>‘x^2-x-2<0$  

$\left(x-2\right)\left(x+1\right)<0$