Answer
$(a) \quad f(3) = 10 $
$(b) \quad f(-x)= 2x^{2}+3x+1 $
$(c)\quad f(x+h) =2x^{2}+(4h-3)x+2h^{2}-3h+1$
Work Step by Step
Given: $f(x)=2 x^{2}-3 x+1$
(a) To find $f(3)$, substitute $x=3$ to obtain:
\begin{align*}
f(3)&=2(3)^{2}-3(3)+1\\
f(3)&=18-9+1\\
f(3)&=10
\end{align*}
(b) To find $f(-x)$, replace $x$ by $-x$:
\begin{align*}f(-x) &=2(-x)^{2}-3(-x)+1 \\
f(-x)&=2 x^{2}+3 x+1 \end{align*}
(c) To find $f(x+h)$, replace $x$ by $(x+h)$ to obtain
\begin{align*}
f(x+h)&= 2(x+h)^{2}-3(x+h)+1 \\
f(x+h)&= 2\left(x^{2}+2 x h+h^{2}\right) -3 x-3 h+1 \\
f(x+h)&=2 x^{2}+4 x h+2 h^{2} -3 x-3 h+1 \\
f(x+h)&=2 x^{2}+(4h-3)x +2 h^{2}-3 h+1 \end{align*}