5. Let $f(x)=a \sin [(x+1) \pi]+b \sqrt[3]{x-1}+2$, where $a$ and $b$ are real constants. If $f(\lg 5)=5$, then $f(\lg 20)=$
5. -1 .

Obviously,
$$
f(x)=a \sin [(x-1) \pi]+b \sqrt[3]{x-1}+2 .
$$

Let $t=x-1$. Then
$$
f(t+1)=a \sin \pi t+b \sqrt[3]{t}+2=g(t)+2,
$$

where, $g(t)=a \sin \pi t+b \sqrt[3]{t}$ is an odd function.
According to the problem,
$$
\begin{aligned}
5 & =f(\lg 5)=f(1-\lg 2) \\
& =g(-\lg 2)+2=-g(\lg 2)+2 .
\end{aligned}
$$

Thus, $g(\lg 2)=-3$.
$$
\text { Therefore, } f(\lg 20)=f(1+\lg 2)=g(\lg 2)+2=-1 \text {. }
$$