So the problem says: Study the bijectivity of the functions f:R->R, which follow the property:
a) $2f(x)+f(1-x)=x-8$; b)$2f(3-2x)+f(\frac{3-x}{2})=x$, where x is a real number for both.
The first one, a), I solved it by substituing $y=1-x=>x=1-y$, gets us the following ecuation:
$2f(1-x)+f(x)=-x-7=>f(1-x)=\frac{-x-7-f(x)}{2}$, and if we substitute this in the first ecuation we get:
$2f(x)+\frac{-x-7-f(x)}{2}=x-8=>4f(x)-f(x)-x-7=2x-16<=>3f(x)=3x-9=>f(x)=x-3$
And this function is indeed bijective. Now, for the second one, b), I don't know what to substitute x with to find the function f. I tried a lot of systems of ecuation with no luck, like $y=\frac{3-x}{2}=>x=3-2y$, which will get us the echivalent functional ecuation: $2f(4y-3)+f(y)=3-2y$, and $y=3-2x=>x=\frac{3-y}{2}$ which will get us to $2f(y)+f(\frac{3-y}{4})=\frac{3-y}{2}$. I don't know what connection to find between these ecuations...
On the second equivalent equation you found,
$2f(y)+f(\frac{3-y}{4})=\frac{3-y}{2}$, (1)
use twice the map $x\mapsto (3-x)/2$, ie, let
$z=\frac{3-(\frac{3-y}{2})}{2}=\frac{y+3}{4}$,
$\Rightarrow y=4z-3$.
Plugging this in (1) yields
$2f(4z-3)+f(-z)=3-2z$
and by subtracting what we found from your first equation we can conclude
$f(z)=f(-z)$.
Hence $f$ is not injective, as it's an even function.