$$
\begin{aligned}
&1+x^4+x^5=\sum_{i=0}^5 a_i(1+x)^i \\
&=a_0+a_1(1+x)^1+a_2(1+x)^2+a_3(1+x)^3 \\
&\quad+a_4(1+x)^4+a_5(1+x)^5 \\
&\Rightarrow 1+x^4+x^5
\end{aligned}
$$
$$
\begin{aligned}
=a_0+a_1(1+x) &+a_2\left(1+2 x+x^2\right)+a_3\left(1+3 x+3 x^2+x^3\right) \\
&+a_4\left(1+4 x+6 x^2+4 x^3+x^4\right) \\
&+a_5\left(1+5 x+10 x^2+10 x^3+5 x^4+x^5\right) \\
\Rightarrow & 1+x^4+x^5 \\
=a_0+a_1+a_1 x+a_2+2 a_2 x+a_2 x^2+a_3+3 a_3 x \\
+3 a_3 x^2+a_3 x^3+a_4+4 a_4 x+6 a_4 x^2+4 a_4 x^3+a_4 x^4+a_5 \\
&+5 a_5 x+10 a_5 x^2+10 a_5 x^3+5 a_5 x^4+a_5 x^5
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow & 1+x^4+x^5 \\
=&\left(a_0+a_1+a_2+a_3+a_4+a_5\right) \\
&+x\left(a_1+2 a_2+3 a_3+4 a_4+5 a_5\right) \\
+x^2\left(a_2+\right.&\left.3 a_3+6 a_4+10 a_5\right)+x^3\left(a_3+4 a_4+10 a_5\right) \\
&+x^4\left(a_4+5 a_5\right)+x^5\left(a_5\right)
\end{aligned}
$$
On comparing the like coefficients, we get
$$
\begin{aligned}
&a_5=1 \\
&a_4+5 a_5=1 \\
&a_3+4 a_4+10 a_5=0
\end{aligned}
$$
and $a_2+3 a_3+6 a_4+10 a_5=0$
from (1) \& (2), we get $a_4=-4 \ldots(5)$
From (1), (3) \& (5), we get $a_3=+6$...(6)
Now, from $(1),(5)$ and $(6)$, we get
$a_2=-4$