5. Given $f(x)=x^{3}+a x^{2}+b x+2$ is symmetric about the point $(2,0)$, then $f(1)=$
Given $f^{\prime}(x)=3 x^{2}+2 a x+b$, let $f^{\prime \prime}(x)=6 x+2 a=0 \Rightarrow x_{0}=-\frac{a}{3}$, $f\left(-\frac{a}{3}\right)=-\frac{a^{3}}{27}+\frac{a^{3}}{9}-\frac{a b}{3}+2$, substituting $(2,0)$ yields $a=-6, b=7$.
Therefore, $f(x)=x^{3}-6 x^{2}+7 x+2 \Rightarrow f(1)=1-6+7+2=4$.