Three. (50 points) Define the function $f: \mathbf{N}_{+} \rightarrow\{0,1\}$, satisfying
$$
\begin{array}{l}
f(1)=f(2)=1, \\
f(n)=1-f(n-1) f\left(\left[\frac{n}{3}\right]\right),
\end{array}
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
For (1) $n=120$; (2) $n=\frac{3^{2012}-1}{2}$; (3) $n=\frac{3^{2012}+1}{2}$, find $f(n)$.
It is known that $f(3)=0$, and for all $k \in \mathbf{N}_{+}$, $f(3 k)=1-f(3 k-1) f(k)$.
Then $f(3 k) f(k)=f(k)+f(3 k)-1$,
$f(3 k+1)=1-f(3 k) f(k)$
$=2-f(k)-f(3 k)$,
$f(3 k+2)=1-f(3 k+1) f(k)$
$=1-f(k)+f(3 k) f(k)=f(3 k)$.
If $f(k)=0$, then
$$
f(3 k)=f(3 k+1)=f(3 k+2)=1 \text {. }
$$

Since $f(3)=0$, we have
$$
f(3 k+3)=1-f(3 k+2) f(k+1)=0 \text {. }
$$

Thus $f\left(\frac{3^{k+1}-3}{2}\right)=f\left(3^{k}+3^{k-1}+\cdots+3\right)=0$.
(1) $f(120)=f\left(\frac{3^{5}-3}{2}\right)=0$;
(2) From $f\left(\frac{3^{2012}-1}{2}-1\right)=f\left(\frac{3^{2012}-3}{2}\right)=0$,

we get $f\left(\frac{3^{2012}-1}{2}\right)=1$;
(3) $f\left(\frac{3^{2012}+1}{2}\right)=f\left(2+\frac{3^{2012}-3}{2}\right)=0$.