Can anybody help with with (3)
My Solution for (1) and (2)
Put $g(x):=f(x)-x$, which is still continuous on $[a,b]$ and differentiable on $(a,b)$. Observe that $x\in [a,b]$ is a fixed point for $f$ iff $g(x)=0$.
Now if $u<v$ are two distinct fixed points we can apply Rolle's theorem (since $g$ is differentiable on $(u,v)\subseteq (a,b)$) and find some $\xi\in (u,v)$ such that $g'(\xi)=0$, i.e. $f'(\xi)=1$, contradiction.
So there is at most one fixed point. The fact that at least one exists is very well-known: if $g(x)\neq 0$ for any $x\in [a,b]$ then $g(x)$ (by continuity) has always the same sign, but $f(a)\ge a$ and $f(b)\le b$, so $g(a)\ge 0$ and $g(b)\le 0$, contradiction. Thus for some $x$ we have $g(x)=0$ and $x$ is the desired fixed point.
Note that $x \le f(x) \le f(c) = c$ for $x \in [a, c]$ and $c = f(c) \le f(x) \le x$ for $x \in [c, b]$ (if this were not the case, apply intermediate value theorem to $f(x) - x$; also, remember that the function is increasing). Therefore, if $x_1 \le c$, then $x_n \le c$ for all $n$, and $(x_n)$ is increasing. By the monotone convergence theorem, $x_n$ converges to some limit $L$. Using the fact that $f$ is continuous, we have $$L = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(L),$$ hence $L$ is a fixed point of $f$, i.e. $L = c$. Similar analysis can be conducted for when $x_1 \ge c$.