A high school calculus exercise on function continuity. 
Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous function and $f(f(x))+3f(x)=12-2x, \hspace{.2cm} \forall x \in \mathbb{R}$. 
$a$. Prove that there is $x_0 \in \mathbb{R}$ such that $f(x_0)=x_0$
 $b.$ Find $f(2)$
 $c.$ If $$\lim_{x \to +\infty}\frac{f(x)}{x}=\lim_{x \to -\infty}\frac{f(x)}{x}=\mathcal{L} \in (-2, 0),$$
Find $\mathcal{L}$

My work so far: 
Suppose that $f(x_1)=f(x_2)\Rightarrow f(f(x_1))=f(f(x_1))\Rightarrow f(f(x_1))+3f(x_1)=f(f(x_2))+3f(x_2)\Rightarrow 12-2x_1=12-2x_2\Rightarrow \\ x_1=x_2$.
So we have concluded that $f$ is one-to-one. Now, for part $a$, suppose that $\nexists \hspace{.1cm} x_0$ such that $f(x_0)=x_0$, so $f(x_0)\neq x_0\hspace{.1cm} (1) \forall x_0 \in \mathbb{R}$ 
$(1)\Rightarrow f(f(x_0))\neq f(x_0)\overset{+3f(x_0)}\Rightarrow 12-2x_0\neq  4f(x_0)\Rightarrow 2f(x_0)+x_0-6\neq 0$. The idea was to suppose some $g(x)=2f(x)+x-6$ and try to check its signs, to show that it may have a solution. Honestly, I'm at a loss. If I could prove $a.$, part $b$ would be so easy. Because then $f(x_0)=x_0$ for some $x_0$ and with the first equation, $x_0=2$. I haven't even tried part $c$, obviously because I can't even do part $a$. The question was taken from a Greek book printed in 1995 and no longer in the market. I doubt it will help listing it.
 A: *

* Suppose that :
$$\forall x \in \mathbb{R}, f(x) > x$$
then :
$$\forall x \in \mathbb{R}, 12 - 2 x = f(f(x)) + 3 f(x) > f(x) + 3 x > x + 3 x = 4 x$$
then :
$$\forall x \in \mathbb{R}, 12 > 6x$$
Absurd.

* Suppose that :
$$\forall x \in \mathbb{R}, f(x) < x$$
then :
$$\forall x \in \mathbb{R}, 12 - 2 x = f(f(x)) + 3 f(x) < f(x) + 3 x < x + 3 x = 4 x$$
then :
$$\forall x \in \mathbb{R}, 12 < 6x$$
Absurd.

We deduce that :
$$\exists a, b \in \mathbb{R}, f(a) \leq a \text{ and } f(b) \geq b$$
then $f(a) - a \leq 0$ and $f(b) - b \geq 0$ and by the intermediate value theorem :
$$\exists x_0 \in \mathbb{R}, f(x_0) - x_0 = 0$$
then $f(x_0) = x_0$.
For the second question :
$$12 - 2 x_0 = f(f(x_0)) + 3 f(x_0) = f(x_0) + 3 x_0 = x_0 + 3 x_0 = 4 x_0$$ 
then $x_0 = 2$ and finally $f(2) = f(x_0) = x_0 = 2$.
For the last question : we have :
$$\lim_{x \to +\infty} \dfrac{f(x)}{x} = \ell < 0$$
then :
$$\lim_{x \to +\infty} f(x) = -\infty$$
and since :
$$\lim_{x \to -\infty} \dfrac{f(x)}{x} = \ell$$
we deduce that :
$$\lim_{x \to -\infty} \dfrac{f(f(x))}{f(x)} = \ell$$
We have :
$$\lim_{x \to +\infty} f(x) = -\infty < 0$$
then :
$$\exists A > 0, \forall x \geq A, f(x) \neq 0$$
then :
$$\forall x \geq A, \dfrac{f(f(x))}{f(x)} \dfrac{f(x)}{x} + 3 \dfrac{f(x)}{x} = \dfrac{12}{x} - 2$$
when $x \to +\infty$ :
$$\ell \, \ell + 3 \ell = -2$$
then $\ell^2 + 3 \ell + 2 = 0$ so $\ell = -2$ or $\ell = -1$.
We deduce that $\ell = -1$ because $\ell \in (-2, 0)$. 
