Let consider the following system :
$\left\{\begin{array}{rl} (x-1)(x-2)(x-4) &\equiv 0 \pmod{9} \\ (x-1)(x-2)(x-4) &\equiv 0 \pmod{25} \\ \end{array} \right.$
We know that $\pmod{9}$ the polynomials has at most $3$ solutions, same for $\pmod{25}$.
The fact is that we don't have $\mathbb{Z}/9\mathbb{Z}$ and $\mathbb{Z}/25\mathbb{Z}$ which are not integral domains (they are not fiels indeed). We can consider $\mathbb{F}_{9}$ and $\mathbb{F}_{25}$ which are fields but this times we cannot apply the CRT.
What methods could be useful ?
Thanks in advance
\begin{align} (x-1)(x-2)(x-4) &\equiv 0 \pmod{9} \\ (x-1)(x-2)(x-4) &\equiv 0 \pmod{25} \\ \end{align}
From the second equation, we know that one possibility is $$x = 25u+1$$ for some integer $u$. Substituting into the first, we get \begin{align} (25u)(25u-1)(25u-3) &\equiv 0 \pmod{9}\\ (7u)(7u-1)(7u-3) &\equiv 0 \pmod{9}\\ 4(7u)\,4(7u-1)\,4(7u-3) &\equiv 0 \pmod{9}\\ u(u-4)(u-3) &\equiv 0 \pmod{9}\\ \hline u &= 9v\\ u &= 9v+4\\ u &= 9v+3\\ \hline x &= 225v + 1 \\ x &= 225v + 101 \\ x &= 225v + 76 \\ \hline x &\equiv 1 \pmod{225}\\ x &\equiv 76 \pmod{225} \\ x &\equiv 101 \pmod{225} \\ \end{align}
etcetera
Note
It's not hard to show that $(x-1)(x-2)(x-4) \equiv 0 \pmod{25} $ only has solutions $\{1,2,4\} \pmod{25}$.
Since $\mathbb Z_5$ is a field, the solutions to $(x-1)(x-2)(x-4) \equiv 0 \pmod{5} $ are $\{1,2,4\} \pmod{5}$
If you try $x = 5t+1$, you get
\begin{align} 5t(5t-1)(5t-3) &\equiv 0 \pmod{25} \\ t(5t-1)(5t-3) &\equiv 0 \pmod 5 \\ t(-1)(-3) &\equiv 0 \pmod 5 \\ 3t &\equiv 0 \pmod 5 \\ t &\equiv 0 \pmod 5 \end{align}
So $x \equiv 1 \pmod{25}$
The same will go for the other two roots.
A SECOND SOLUTION
Let's start with $(x-1)(x-2)(x-4) \equiv 0 \pmod 9$ This implies, but is not equivalent to, the simpler equation $(x-1)(x-2)(x-4) \equiv 0 \pmod 3$. We can simplify this to $(x-1)^2(x-2) \equiv 0 \pmod 3$, and, because $\mathbb Z_3$ is a field, the solution set is $x \in \{1,2\} \pmod 3$.
If $x \equiv 1 \pmod 3$, Then $x = 3t+1$ for some integer $t$. Then
\begin{align} (x-1)(x-2)(x-4) &\equiv 0 \pmod 9 \\ (3t)(3t-1)(3t-3) &\equiv 0 \pmod 9 \\ 9t(3t-1)(t-1) &\equiv 0 \pmod 9 \\ 0 &\equiv 0 \pmod 9 \end{align}
So $t \in \mathbb Z$. This implies that $x \in \{\dots, -5, -2, 1, 4, 7, \dots\}$, which simplifies to $x \in \{1, 4, 7\} \pmod 9$
If $x \equiv 2 \pmod 3$, Then $x = 3t+2$ for some integer $t$. Then
\begin{align} (x-1)(x-2)(x-4) &\equiv 0 \pmod 9 \\ (3t+1)(3t)(3t-2) &\equiv 0 \pmod 9 \\ 3t(3t+1)(3t-2) &\equiv 0 \pmod 9 \\ t(3t+1)(3t-2) &\equiv 0 \pmod 3 \\ t(1)(-2) &\equiv 0 \pmod 3 \\ -2t &\equiv 0 \pmod 3 \\ t &\equiv 0 \pmod 3 \\ \end{align}
So $x \in \{\dots, -16, -7, 2, 11, 20, \dots\}$. This implies that $x \in \{2\} \pmod 9$.
Putting the two solutions together, we get $x \in \{1,2,4,7\} \pmod 9$.
So now we tackle $(x-1)(x-2)(x-4) \equiv 0 \pmod{25}$. Just as before, this implies that $(x-1)(x-2)(x-4) \equiv 0 \pmod 5$, which implies that $x \in \{1,2,4\} \pmod 5$.
If $x \equiv 1 \pmod 5$, Then $x = 5t+1$ for some integer $t$. Then
\begin{align} (x-1)(x-2)(x-4) &\equiv 0 \pmod{25} \\ (5t)(5t-1)(5t-3) &\equiv 0 \pmod 5 \\ t(5t-1)(5t-3) &\equiv 0 \pmod 5 \\ 3t &\equiv 0 \pmod 5 \\ t \equiv 0 \pmod 5 \\ \end{align}
Which implies that $x \in \{1\} \pmod{25}$.
The other two cases are similar; so we can conclude that $x \in \{1,2,4\} \pmod{25}$.
By the Chinese Remainder Theorem, $\bar a \in \mathbb Z_9$ is a solution to $(x-1)(x-2)(x-4) \equiv 0 \pmod 9$ and $\bar b \in \mathbb Z_{25}$ is a solution to $(x-1)(x-2)(x-4) \equiv 0 \pmod{25}$ if and only if $\bar c = 100 \bar a - 99 \bar b \in \mathbb Z_{225}$ is a solution to $(x-1)(x-2)(x-4) \equiv 0 \pmod{225}$.
We can make a table.
$\begin{array}{cc|c} a \pmod 9 & b\pmod{25} & 100a - 99b \pmod{225} \\ \hline 1 & 1 & 1 \\ 2 & 1 & 101 \\ 4 & 1 & 76 \\ 7 & 1 & 151 \\ 1 & 2 & 127 \\ 2 & 2 & 2\\ 4 & 2 & 202 \\ 7 & 2 & 52 \\ 1 & 4 & 154 \\ 2 & 4 & 29 \\ 4 & 4 & 4 \\ 7 & 4 & 79 \\ \hline \end{array}$
So the solution set is
$x \in \{1, 2, 4, 29, 52, 76, 79, 101, 127, 151, 154, 202\} \pmod{225}$