The answer to both questions 1 and 2 is false, due to the following example. Consider $(n+1)$ consecutive intervals $I_0,...,I_n$ of lengths $1,2,4,...,2^{n-1}, 2^n$. Let the map $T_n$ cyclically permute them in an affine way. In fact, if you choose $I_j=[2^j,2^{j+1}]$, then $$ T_{n+1}(x)=\begin{cases} 2x, & x \notin I_{n}\\ 2^{-n}x, & x\in I_{n}. \end{cases} $$
Now, make an affine change of the coordinates so that the union $I_0\cup\dots\cup I_n=[1,2^{n+1}]$ becomes (always the same) circle $[0,1]/(0\sim 1)$. Then, the constructed maps on this circle $C^0$-converge, as $n\to\infty$, to $$ T(x)=\begin{cases} 2x, & x \in [0,1/2]\\ 0, & x\in [1/2,1]. \end{cases} $$
This map is not a homeomorphism (it collapses the interval $[1/2,1]$ to a single point 0). Also, the orbit of any point falls to the fixed point 0 in a finite number of steps, so there is only one periodic point, and there is no dense orbit.