Given quadratic equation $2x^2+bx+\frac{1}{b}=0,$ has two distinct real

roots, so

$\Rightarrow$ $\begin{array}{r}D>0\\ b^2-4\left(2\right)\left(\frac{1}{b}\right)>0\end{array}$

$\Rightarrow b^2-\frac{8}{b}>0\Rightarrow \frac{b^3-8}{b}>0$

$\Rightarrow \frac{(b-2)\left(b^2+2b+4\right)}{b}>0$

$\Rightarrow$ $b\in (-\infty ,0)\cup (2,\infty )$

For option (c), $b^2-3b>-2$

$\Rightarrow b^2-3b+2>0$

$\Rightarrow$ $(b-2)(b-1)>0$

$b\in (-\infty ,1)\cup (2,\infty )$

mean if $b\in (-\infty ,0)\cup (2,\infty )$

then $b^2-3b>-2$