Let $B_{n}$ be the event that there are $k$ heads starting from toss $n$ . Then, $A_{k}=\cup_{n=2^{k}}^{2^{k+1}-k}B_{n}$ , and $$\begin{eqnarray*} \mathbb{P}(A_{k}) & \le & \sum_{n=2^{k}}^{2^{k+1}-k}\mathbb{P}(B_{n})\\ & \le & 2^{k}p^{k}, \end{eqnarray*}$$ so that for $p<\tfrac{1}{2}$ , $$\begin{eqnarray*} \sum_{k\in\mathbb{N}}\mathbb{P}(A_{k}) & \le & \tfrac{2p}{1-2p}\\ & < & \infty, \end{eqnarray*}$$ and, by BC1, $\mathbb{P}(A_{k},\mbox{ i.o.})=0$ .

Regarding the second part of the statement, I am not sure how we can use the inclusion-exclusion formula, especially because if we consider it for events $E_{i}$ , the sum of the first two terms to estimate the probability from below becomes negative for large $k$ . However, consider events $E_{i}^{k}$ for $k\in\mathbb{N}$ and $i\in\{1,\left\lfloor \tfrac{2^{k}}{k}\right\rfloor \}$ . These events are independent, $\{E_{i}^{K},\mbox{ i.o.}\}$ implies $\{A_{k},\mbox{ i.o.}\}$ . Further, for $p\ge\tfrac{1}{2}$ , $$\begin{eqnarray*} \sum_{k\in\mathbb{N}}\sum_{i=1}^{\left\lfloor \tfrac{2^{k}}{k}\right\rfloor }\mathbb{P}(E_{i}^{k}) & \ge & \sum_{k\in\mathbb{N}}\left(\tfrac{2^{k}}{k}-1\right)p^{k}\\ & \ge & \sum_{k\in\mathbb{N}}\tfrac{1}{k}-\tfrac{p}{1-p}\\ & = & +\infty, \end{eqnarray*}$$ hence, by BC2, $\mathbb{P}(E_{i}^{K},\mbox{ i.o.})=1$ , and $\mathbb{P}(A_{k},\mbox{ i.o.})=1$ .