$\begin{align}  & a)A=\{x:x\in \mathbb{N},{{x}^{2}}-5x+6=0\} \\ & {{x}^{2}}-5x+6=0 \\ & (x-2)(x-3)=0 \\ & x=2\text{ }or\text{ }x=3 \\ & 2\text{ and 3 are natural numbers, so they member of the set }A \\ & A=\{2,3\} \\ & b)B=\{y:y\in \mathbb{Z},2{{y}^{2}}-3y+1=0 \\ & 2{{y}^{2}}-3y+1=0 \\ & 2{{y}^{2}}-2y-y+1=0 \\ & 2y(y-1)-1(y-1)=0 \\ & (2y-1)(y-1)=0 \\ & y=\tfrac{1}{2}\text{ or }y=1 \\ & \text{we shall ignore }\tfrac{1}{2}\text{ because it not an integer and pick 1} \\ & B=\{1\} \\ & c)C=\{x:x\in \mathbb{Q};6{{x}^{2}}-13x+6=0\} \\ & 6{{x}^{2}}-13x+6=0 \\ & 6{{x}^{2}}-4x-9x+6=0 \\ & (2x-3)(3x-2)=0 \\ & x=\tfrac{3}{2}\text{ or }x=\tfrac{2}{3} \\ & \text{Both }\tfrac{3}{2}\text{ and }\tfrac{2}{3}\,\text{are rational numbers} \\ & \therefore x=\{\tfrac{2}{3},\tfrac{3}{2}\} \\ & d)D=\{x:x\in \mathbb{Q};{{x}^{2}}-x+1=0\} \\ & {{x}^{2}}-x+1=0 \\ & \text{using the quadratic formula} \\ & x=\frac{\text{1}\pm \sqrt{{{1}^{2}}-4\cdot 1\cdot 1}}{2\cdot 1}=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm \sqrt{-3}}{2} \\ & \text{the resulting solution is a complex, so the set }D \\ & \text{is a null set} \\ & D=\{\}\text{ or }\varnothing  \\ & e)\text{the set of all integers whose squares are less than 20} \\ & \text{Let the set be }A \\ & A=\{-1,-2,-3,-4,0,1,2,3,4\} \\ & f)\text{the set of all integers between }-3\text{ and }+3 \\ & \text{Let the set be }B \\ & B=\{-2,-1,0,1,2,\} \\ & g)P=\{x\in \mathbb{N}:-1\le x\le 1\} \\ & P=\{\}\text{ or }\varnothing  \\\end{align}$