How to prove these four propostions about Borel-Cantelli Lemma in Shriyaev's book 'Probability'? The original problems are shown in picture linked below. Thank you.

Problem $\mathbf{2.10.19}$. (On the second Borel-Cantelli lemma.)  Prove the following variants of the second Borel-Cantelli lemma: given an arbitrary sequence of (not necessarily independent) events $A_1, A_2, \ldots$, one can claim that:
(a) If
$$
\sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad
\liminf_n \frac{\sum_{i,k=1}^n {\sf P}(A_iA_k)}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} = 1,
$$
then (Erdös and Rényi [$37$]) ${\sf P}(A_n\text{ i.o.}) = 1$.
(b) If
$$
\sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad
\liminf_n \frac{\sum_{i,k=1}^n {\sf P}(A_iA_k)}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} = L,
$$
then (Kochen and Stone[$64$], Spitser [$125$]) $L \geq 1$ and ${\sf P}(A_n\text{ i.o.}) = 1/L$.
(c) If
$$
\sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad
\liminf_n \frac{\sum_{1\leq i<k\leq n} [{\sf P}(A_iA_k)-{\sf P}(A_i){\sf P}(A_k)]}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} \leq 0,
$$
then (Ortega and Wschebor [$92$]) ${\sf P}(A_n\text{ i.o.}) = 1$.
(d) If $\sum_{n=1}^\infty {\sf P}(A_n) = \infty$ and
$$
\alpha_H = \liminf_n \frac{\sum_{1\leq i<k\leq n} [{\sf P}(A_iA_k)-H{\sf P}(A_i){\sf P}(A_k)]}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2},
$$
where $H$ is an arbitrary constant, then (Petrov [$95$]) ${\sf P}(A_n\text{ i.o.}) \geq \frac{1}{H+2\alpha_H}$ and $H+2\alpha_H \geq 1$.

Original at https://i.stack.imgur.com/oxcKz.jpg
 A: This is the Kochen-Stone Lemma. I will state this result and a short proof for you. But first a little technical result.
Lemma: If $0\neq f\in L_2$ and $\mathbb{E}[f]\geq0$,  then for any $0<\lambda<1$
$$\begin{align}
\mathbb{P}\big[f>\lambda \mathbb{E}[f]\big]\geq (1-\lambda)^2 \frac{\big(\mathbb{E}[f]\big)^2}{\mathbb{E}[|f|^2]}\tag{1}\label{anty-cheby}.
\end{align}
$$
Here is a short proof:
By Hölder's inequality
$$
\mathbb{E}[f]=\int_{\{f\leq \lambda\mathbb{E}[f]\}}f \,d\Pr+ \int_{\{ f>\lambda\mathbb{E}[f]\}} f\,d\mathbb{P} \leq \lambda\mathbb{E}[f] + \Big(\|f\|_2\sqrt{\Pr[f>\lambda\mathbb{E}[f]]}\Big). 
$$
Here is the result that we will used to get the version of Corel Cantelly closer to what you described in your problem.
Lemma(Kochen-Stone) Let $\{A_n\}\subset\mathscr{F}$. If $\sum_n\mathbb{P}[A_n]=\infty$, then
$$\begin{align}
\mathbb{P}\big[\bigcap_{n\geq1}\bigcup_{k\geq n}A_k\big]\geq\limsup_n\frac{\Big(\sum^n_{k=1}\mathbb{P}[A_k]\Big)^2}{\sum^n_{k=1}\sum^n_{m=1}\mathbb{P}[A_k\cap A_m]}\tag{2}\label{ko-sto}
\end{align}
$$
Here is a Sketch of the proof:
Without loss of generality, we assume that $\mathbb{P}[A_n]>0$ for all $n$. Let $f_n=\sum^n_{k=1}\mathbb{1}_{A_k}$, $f=\sum_{n\geq1}\mathbb{1}_{A_n}$, and for any $0<\lambda<1$, define  $B_{n,\lambda}=\big\{f_n>\lambda\mathbb{P}[f_n]\big\}$.
Observe that
$$
A=\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k=\{f=\infty\}\supset\bigcap_{n\geq 1}\bigcup_{k\geq n}B_{k,\lambda}=B_\lambda;
$$
then,  by $\eqref{anty-cheby}$, we obtain
$$
\mathbb{P}[A]\geq\mathbb{P}[B_\lambda]\geq\limsup_{n\rightarrow\infty}\mathbb{P}[B_{n,\lambda}]\geq(1-\lambda)^2\limsup_n\frac{\big(\mathbb{E}[f_n]\big)^2}{\mathbb{E}[f^2_n]}.
$$
Letting $\lambda\rightarrow1$ gives $\eqref{ko-sto}$.
Using Kochen-Stone's Lemma one can prove the following version of the reverse Borel-Cantelli Lemma
Theorem (reverse Borel-Cantelli) Suppose $\{A_n\}\subset\mathscr{F}$ is such that  for any $i\neq j$, $\mathbb{P}[A_i\cap  A_j]\leq\mathbb{P}[A_i]\mathbb{P}[A_j]$. If  $\sum_n\mathbb{P}[A_n]=\infty$,  then $\mathbb{P}\Big[\bigcap_{n\geq1}\bigcup_{k\geq n}A_k\Big]=1$.
Here is a short proof:
Denote by  $A=\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k$. Let  $a_n=\sum^n_{k=1}\mathbb{P}[A_k]$,, $b_n=\sum_{i\neq j}\mathbb{P}[A_i]\mathbb{P}[ A_j]$, and $c_n=\sum^n_{k=1}\mathbb{P}^2[A_k]$. By Kochen--Stone's lemma we have
$$
\mathbb{P}[A]\geq\limsup_n\frac{c_n+b_n}{a_n+b_n}
$$
From $a^2_n=c_n+b_n\leq a_n+b_n$, and  $a_n\nearrow\infty$, it follows that $b_n\nearrow\infty$ and  $\lim_n\tfrac{c_n}{b_n}=0=\lim_n\frac{a_n}{b_n}$. Therefore,  $\mathbb{P}[A]=1$.

Reference:
https://projecteuclid.org/euclid.ijm/1256059668
