$\text{a)}{\left(3x-2\right)}^2-5x-\left(3x+2\right)\left(3x-2\right)<{\left(2-3x\right)}^2-9x^2+4$  

$9x^2-\cancel{12x}+\cancel{4}-5x-\left(9x^2-4\right)<4-\cancel{12x}+\cancel{9x^2}-\cancel{9x^2}+\cancel{4}$  

$\cancel{9x^2}-5x-\cancel{9x^2}+\cancel{4}<\cancel{4}$  

$-5x<0\mid :\left(-5\right)$  

$x>0$  

$x\in \left(0,+\infty \right)$  

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$\text{b)}2{\left(x-3\right)}^2-\frac{1}{2}\ge \frac{\left(2x-3\right)\left(2x+3\right)}{2}-x$  

$2\left(x^2-6x+9\right)-\frac{1}{2}\ge \frac{4x^2-9}{2}-x\mid \cdot 2$  

$4\left(x^2-6x+9\right)-1\ge 4x^2-9-2x$  

$\cancel{4x^2}-24x+36-1\ge \cancel{4x^2}-2x-9$  

$-22x\ge -44\mid :\left(-22\right)$  

$x\le 2$  

$x\in (-\infty ,2⟩$  

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$\text{c)}{\left(x-3\right)}^2-\frac{x^2-3}{3}-9<\frac{\left(x-2\right)\left(x+2\right)}{6}+\frac{1}{2}{x}^2$  

$x^2-6x+\cancel{9}-\frac{x^2-3}{3}-\cancel{9}<\frac{x^2-4}{6}+\frac{1}{2}{x}^2$  

$x^2-6x-\frac{x^2-3}{3}<\frac{x^2-4}{6}+\frac{1}{2}{x}^2\mid \cdot 6$  

$6x^2-36x-2\left(x^2-3\right)<x^2-4+3x^2$  

$6x^2-36x-2x^2+6<4x^2-4$  

$\cancel{4x^2}-36x+6<\cancel{4x^2}-4$  

$-36x<-10\mid :\left(-36\right)$  

$x>\frac{5}{18}$  

$x\in \left(\frac{5}{18},+\infty \right)$  

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$\text{d)}1-{\left(3x-2\right)}^2\ge \frac{x-36}{3}-9\left(x+1\right)\left(x-1\right)$  

$1-\left(9x^2-12x+4\right)\ge \frac{x-36}{3}-9\left(x^2-1\right)$  

$1-\cancel{9x^2}+12x-4\ge \frac{x-36}{3}-\cancel{9x^2}+9$  

$12x-3\ge \frac{x-36}{3}+9\mid \cdot 3$  

$36x-9\ge x-36+27$  

$36x-\cancel{9}\ge x-\cancel{9}$  

$35x\ge 0\mid :35$  

$x\ge 0$  

$x\in ⟨0,+\infty )$