This question has a lot of parts so I'll post each part separately.
First, show $f(x) = x$ in $(\mathbb{Z}/6\mathbb{Z})[x]$ factors as $(3x+4)(4x+3)$
I am trying long division. I cannot divide $x$ by either $(3x+4)$ or by $(4x+3)$ mod 6. Anything multiplied by 3 is either 3 or 0 mod 6. And 4,2,0 mod 6 when a number is multiplied by 4.
HINT
Note that $$ (3x+4)(4x+3) = 12x^2 + 25x + 12 \equiv x \pmod{6} $$
More conceptually, note that $\,\begin{align}f \equiv 3\\ e\equiv 4\end{align}\,$ obey $\,\begin{align}e^2\equiv&\ e,\ \ e\!+\!f \equiv 1\\ f^2\equiv&\ f,\ \ \ \ \color{#c00}{e\,f\,\equiv 0}\end{align}\ \ $ [orthogonal idempotents], so
$$ (ex\!+\!f)(fx\!+\!e)\, \equiv\, \color{#c00}{ef}\, x^2 + (e^2\!+\!f^2) x + \color{#c00}{ef}\,\equiv\, x\qquad\qquad$$
The same will occur with the idempotents arising from any CRT direct product decomposition.
Essentially what occurs is that the trivial factorization $x \equiv x\cdot 1\,$ becomes nontrivial by permuting the factors $\,x,1\,$ in each product component, as below
$$\underset{}{\overset{\Large \bmod 2:}{\phantom{I^{I^{I^{I^{I^I}}}}}}}\underbrace{\overbrace{(3x\!+\!4)}^{\Large \equiv\ x }}_{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Large\bmod 3:\ \ \ \ \ \equiv\ 1}\,\underbrace{\overbrace{(4x\!+\!3)}^{\Large \equiv\ 1}}_{\Large \equiv\ x}\qquad $$
where we have that $\, x\equiv gh \equiv x\cdot 1\pmod{\!2}\,$ vs. $\ gh\equiv 1\cdot x\pmod{\!3}$
