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 
$$
T_n(x)=\begin{cases}
2x, & x \not\in I_{{n}}\\
2^{-n}x, & x\in I_{n}.
\end{cases}
$$

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.