$a\text{)}{\left(x+3\right)}^2=\left(x-2\right)\left(x+2\right)$  

$x^2+6x+9=x^2-4\mid -x^2$      

$6x+9=-4$  

$6x=-13$  

$x=-\frac{13}{6}$  

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$b\text{)}\frac{4x-5}{2x-1}=\frac{2}{3}$  

założenia:

$2x-1\ne 0$  

$2x\ne 1$  

$x\ne \frac{1}{2}$  

zatem:

$3\cdot \left(4x-5\right)=2\cdot \left(2x-1\right)$  

$12x-15=4x-2$  

$8x=13$  

$x=\frac{13}{8}$