2. Given: $b_{1}, b_{2}, b_{3}, b_{4}$ are positive integers, the polynomial $g(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}}\left(1-z^{4}\right)^{b_{4}}$ when expanded and terms higher than 4th degree are omitted, becomes $1-2 z$. Also, $\alpha$ is the largest root of the polynomial $f(x)=x^{3}-b_{4} x^{2}+b_{2}$. Try to find the remainder when $\left[\alpha^{1995}\right]$ is divided by 9. Here $[x]$ denotes the greatest integer less than or equal to $x$.
2. $g(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}}\left(1-z^{4}\right)^{b_{4}}$ Comparing the coefficients of $z, z^{2}, z^{3}, z^{4}$, we have:
$$
\begin{array}{l}
b_{1}=2, b_{2}=1, b_{3}=2, b_{4}=3, \\
\therefore \quad f(x)=x^{3}-3 x^{2}+1 .
\end{array}
$$

Let the three roots of $f(x)=0$ be $\alpha, \beta, \gamma$, and assume $\alpha>\beta>\gamma$,
$$
\begin{array}{l}
f(-1)=-30, f(1)0, f\left(\frac{\sqrt{2}}{2}\right)\beta>0 .
\end{array}
$$

By Vieta's formulas, $\alpha+\beta+\gamma=3 \Rightarrow \beta+\gamma=3-\alpha>0$,
$$
\begin{array}{l}
\therefore \quad|\beta|>|\gamma| . \\
\therefore \quad \frac{\sqrt{2}}{2}>\beta>-\gamma>0 . \\
\therefore \quad \beta^{1995}+\gamma^{1995}=|\beta|^{1995}-|\gamma|^{1995}>0, \\
\beta^{1995}+\gamma^{1995}=|\beta|^{1995}-|\gamma|^{1995}<1 . \\
\therefore \quad \alpha^{1995}<\alpha^{1995}+\beta^{1995}+\gamma^{1995}<\alpha^{1995}+1 . \\
\therefore \quad \text { Let } A_{n}=\alpha^{n}+\beta^{n}+\gamma_{n} .
\end{array}
$$

Then $A_{n+3}=3 A_{n+2}-A_{n}$.
$$
A_{0}=3, A_{1}=3, A_{2}=9 \text {. }
$$
$\therefore A_{n} \in \mathbf{Z}, n \geqslant 0$, from (1) and this, we know
$\left[\alpha^{1995}\right]=A_{1995}-1$ Now find $A_{1995}$ modulo 9
Let $a_{k} \equiv A_{k}(\bmod 9)$,
$0 \leqslant a_{k} \leqslant 8$, then $a_{0}=3, a_{1}=3, a_{2}=0, a_{3}=6, a_{4}=6, a_{5}=0, a_{6}=3, a_{1}=3$.
Next, we prove $a_{k}=a_{k+6}, k=0,1,2,3$; the conclusion is obviously true for $k=0,1,2,3$. Assume the conclusion holds for $k \leqslant k$, then $a_{k+1} \equiv 3 a_{k}-a_{k-2} \equiv 3 a_{k+6}-a_{k+2} \equiv a_{k+7}(\bmod 9)$,
$\therefore$ the conclusion also holds for $k+1$, $\therefore \quad a_{k} \equiv a_{k+6}(\bmod 9)$.
$\therefore a_{1995} \equiv a_{1995-6} \equiv \cdots \equiv a_{3} \equiv 6(\bmod 9)$.
$\therefore \quad\left[\alpha^{1995}\right] \equiv A_{1995}-1 \equiv 5(\bmod 9)$.
$\therefore\left[\alpha^{1995}\right]$ modulo 9 is 5.