5. Given $f(x)=x^{3}+a x^{2}+b x+2$ is symmetric about the point $(2,0)$. Then $f(1)=$ $\qquad$
5. 4 .

From the symmetry of the graph of $f(x)$ about the point $(2,0)$, we know
$$
\begin{array}{c}
f(x+2)=(x+2)^{3}+a(x+2)^{2}+b(x+2)+2 \\
=x^{3}+(a+6) x^{2}+(b+4 a+12) x+ \\
4 a+2 b+10
\end{array}
$$

is an odd function.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
a+6=0, \\
4 a+2 b+10=0
\end{array}\right. \\
\Rightarrow(a, b)=(-6,7) \\
\Rightarrow f(1)=1+a+b+2=4 .
\end{array}
$$