A Non-Decreasing Function

A function $f(x)$ on an interval $D$ is non-decreasing if $f(x_1) \leqslant f(x_2)$ whenever $x_1 < x_2$ in $D$. Suppose $f(x)$ is non-decreasing on $[0, 2]$ with

and $f(x) \leqslant 2(x - 1)$ for all $x \in \left[\dfrac{3}{2}, 2\right]$. Determine, with proof, which of the following are true:

1. $f(1) = 1$; 2. there exists $x_0 \in \left[\dfrac{3}{2}, 2\right]$ with $f(x_0) < 1$; 3. $f\!\left(\dfrac{1}{4}\right) + f\!\left(\dfrac{2}{3}\right) + f\!\left(\dfrac{25}{18}\right) + f\!\left(\dfrac{7}{4}\right) = 4$; 4. for every $x \in \left[0, \dfrac{1}{2}\right]$, $f(f(x)) \leqslant -f(x) + 2$.