$a\in(0,1)$ is fixed.
$M\in\Bbb Z_{>1}$ is fixed.
What is $f(x)$ given that $$f(0)=0\mbox{, }f(M)=1+a\mbox{, }f(1)=1-a$$$$\mbox{ }f(x)\in(1-a,1+a)\mbox{, }\forall x\in(1,M)?$$
What is $g(x)$ given that $$g(1)=1\mbox{, }g(0)=-a\mbox{, }g(1-M)=a$$$$\mbox{ }f(x)\in(-a,+a)\mbox{, }\forall x\in(1-M,0)?$$
What is $\mathsf{deg}(f(x)),\mathsf{deg}(g(x))$?
This is not a complete answer, but some observations.
For the first part, note that $f(x)$ is of the form $f(x)=xh(x)$. Now, $f(x)$ cannot be $0$ for any real $x$ for $x\in (1,M)$. So the degree of $f$ is of the form $2m+n+1$ where $m$ is the number of complex roots of $f$, $n$ is the number of real roots of $f$, if any, in $(0,1)$.