1. Analyze the limit: ${\lim }_{x\to 0}(1+\frac{f(x)}{x^2})=3$ implies ${\lim }_{x\to 0}\frac{f(x)}{x^2}=2$. For this limit to exist and be finite, $f(x)$ must have 0 coefficients for $x^0$ and $x^1$ terms, and the coefficient of $x^2$ must be 2. Since $f(x)$ is degree 4, let $f(x)=a{x}^4+b{x}^3+2x^2$.

2. Use derivative conditions: $f(x)$ has extremes at $x=1$ and $x=2$, so $f^{\prime }(1)=0$ and $f^{\prime }(2)=0$. $f^{\prime }(x)=4a{x}^3+3b{x}^2+4x=x(4a{x}^2+3bx+4)$.

At $x=1$: $4a+3b+4=0$ (Eq 1) At $x=2$: $32a+12b+8=0⟹8a+3b+2=0$ (Eq 2)

3. Solve for a and b: Subtract Eq 1 from Eq 2: $(8a+3b+2)-(4a+3b+4)=0⟹4a-2=0⟹a=1/2$. From $4a+3b=-4$: $2+3b=-4⟹3b=-6⟹b=-2$.

So $f(x)=\frac{1}{2}{x}^4-2x^3+2x^2$.

4. Calculate $f(2)$: $f(2)=\frac{1}{2}(16)-2(8)+2(4)=8-16+8=0$.