$\because f(x)$ has extremum values at $x=1$ and $x=2$
$\because f^{\prime}(1)=0$ and $f^{\prime}(2)=0$
As, $f(x)$ is a polynomial of degree 4 .
Suppose $f(x)=A x^4+B x^3+C x^2+D x+E$
$$
\begin{aligned}
&\because \lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3 \\
&\Rightarrow \lim _{x \rightarrow 0}\left(\frac{A x^4+B x^3+C x^2+D x+E}{x^2}+1\right)=3 \\
&\Rightarrow \lim _{x \rightarrow 0}\left(A x^2+B x+C+\frac{D}{x}+\frac{E}{x^2}+1\right)=3
\end{aligned}
$$
As limit has finite value, so $D=0$ and $E=0$
Now $A(0)^2+B(0)+C+0+0+1=3$
$$
\begin{aligned}
&\Rightarrow c+1=3 \Rightarrow c=2 \\
&f^{\prime}(x)=4 A x^3+3 B x^2+2 C x+D \\
&f^{\prime}(1)=0 \Rightarrow 4 A(1)+3 B(1)+2 C(1)+D=0
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow 4 A+3 B=-4 \\
f^{\prime}(2)=0 & \Rightarrow 4 A(8)+3 B(4)+2 C(2)+D=0 \\
& \Rightarrow 8 A+3 B=-2
\end{aligned}
$$
From equations (1) and (2), we get $A=\frac{1}{2}$ and $B=-2$
So, $f(x)=\frac{x^4}{2}-2 x^3+2 x^2$
Therefore, $f(-1)=\frac{(-1)^4}{2}-2(-1)^3+2(-1)^2$
$$
=\frac{1}{2}+2+2=\frac{9}{2}
$$
Hence $f(-1)=\frac{9}{2}$