Given,
$$
g(x)=x^2+x-2
$$
and
$$
\begin{aligned}
& \frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2 \\
& \Rightarrow \quad g(f(x))=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+(f(x))-2=4 x^2-10 x+4 \\
&
\end{aligned}
$$
Now, it is necessary that $f(x)$ should be linear polynomial expression, so let $f(x)=a x+b$, then
$$
\begin{array}{r}
(a x+b)^2+(a x+b)-2=4 x^2-10 x+4 \\
\Rightarrow \quad a^2 x^2+(2 a b+a) x+\left(b^2+b-2\right) \\
=4 x^2-10 x+4
\end{array}
$$
On comparing the coefficient of different kind of terms, we are getting
$$
\begin{aligned}
& a^2=4 \text {, } \\
& \Rightarrow \quad 2 a b+a=-10 \\
& \text { and } \quad b^2+b-2=4 \\
&
\end{aligned}
$$
So,
then
$$
\begin{aligned}
& a= \pm 2, \\
& b=\left\{\begin{aligned}
-3 ; & \text { if } a=2 \\
2 ; & \text { if } a-2^{\prime}
\end{aligned}\right.
\end{aligned}
$$
and these value satisfy the all above relations, so
$$
f(x)=2 x-3 \text { or }-2 x+2
$$