A coin tossing challenge!

$\text{Consider a win on the nth roll.}\\ \text{This roll sequence looks like }\\ \underset{n_T}{\underbrace{TTT\dots T}}\underset{n_H}{\underbrace{HHH\dots H}}T\\ 0\leq n_T \leq n-2,~1\leq n_H \leq n-1,~n_T+n_H +1 = n$

$\text{Thus there are }n-1 \text{ ways to accomplish a win on the nth roll}\\ \text{Each sequence has probability }p_n = 2^n \text{thus the probability of a win on the nth roll is given by}\\ P[n] = \dfrac{n-1}{2^n},~~n\in \mathbb{N},~2 \leq n$

$E[N] = \sum \limits_{n=2}^\infty~\dfrac{n(n-1)}{2^n}=6-2= 4$

You can apply the same idea to solve 2 and 3