For 2.
Claim: $\lim_{n\rightarrow\infty}2^{-n/2}\mathbb{E}[K_n]=0$
The idea is this: The union of the events $E_{jn}$ is the event \begin{equation} F_n=\left\{B_t=0 \text{ for some $t\in\left[\frac{2^n+1}{2^n},\frac{2^{2^n}}{2^n}\right]$} \right\} \end{equation}\begin{equation} F_n=\left\{B_t=0 \text{ for some $t\in\left[\frac{2^n+1}{2^n},\frac{2^{2n}}{2^n}\right]$} \right\} \end{equation} with probability less or equal than 1. The intersection of events $E_{jn}$ and $E_{j+1,n}$ is \begin{equation} G_{jn}=\left\{B_{j/2^n}=0\right\} \end{equation} which has probability equal to zero for all $j,n$. Then \begin{eqnarray} \mathbb{E}[K_n] & = & \mathbb{E}\left[\sum_{j=2^n+1}^{2^{2^n}}1_{E_{jn}}\right]\\ & = & \sum_{j=2^n+1}^{2^{2^n}}\mathbb{E}[1_{E_{jn}}]\\ & = & \sum_{j=2^n+1}^{2^{2^n}}\mathbb{P}[E_{jn}]\\ & = & \mathbb{P}[F_{n}]+\sum_{j=2^n+1}^{2^{2^n}}\mathbb{P}[G_{jn}]\\ & \leq & 1 \end{eqnarray}\begin{eqnarray} \mathbb{E}[K_n] & = & \mathbb{E}\left[\sum_{j=2^n+1}^{2^{2n}}1_{E_{jn}}\right]\\ & = & \sum_{j=2^n+1}^{2^{2n}}\mathbb{E}[1_{E_{jn}}]\\ & = & \sum_{j=2^n+1}^{2^{2n}}\mathbb{P}[E_{jn}]\\ & = & \mathbb{P}[F_{n}]+\sum_{j=2^n+1}^{2^{2n}}\mathbb{P}[G_{jn}]\\ & \leq & 1 \end{eqnarray}
This implies $\lim_{n\rightarrow\infty}2^{-n/2}\mathbb{E}[K_n]=0$