Wolstenholme's theorem is stated as follows:
if $p>3$ is a prime, then
\begin{align*}
\sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\
\sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}.
\end{align*}
It is also not hard to prove that
$$
\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}.
$$
However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

**Question:**
If $p$ is an odd prime, then
$$
4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}.
$$
I have verified this congruence for $p$ upto $7919$.

**Comments:** 

(1) Since 
$$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$
we need to prove
$$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$
This idea was given by Fedor Petrov but there are something wrong in his answer. **This question is still open**.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e.,
$$
4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}.
$$
This was conjectured by tkr.

I appreciate any proofs, hints, or references!