On an Elliptic Curve is that possible that from $P$ we can tell if $a$ is quadratic residue modulo $N$?

Question

Imagine that, On an Elliptic Curve cryptography scheme where $P=a\times G$, Bob shares his public key $P$ with Eve (the devil who wants to know secrets he is not supposed to). Bob has also revealed a clue about $a$ accidentally. The clue can be one or a combination of items from the following list:

1. The number $a$ is ODD/EVEN integer.
2. The number $a$ is GREATER/SMALLER than half of the group order $N/2$.
3. The number $a$ has $x$ meaningful bits when written as binary. (there $x$ is the count of bits of $a$, for example if $a=152=10011000$ then $x=8$
4. The number $a$ is quadratic RESIDUE/NON-RESIDUE modulo $N$.

Question 1:

Will knowledge of such a clue be considered a significant weakness for the public key of Bob so that we should say it is not safe anymore to use it?

Question 2:

Clues mentioned above are very little information about $a$ I suppose. Am I right? What if we can reveal them for all points on the curve by using a magic algorithm?

I know that, for items 1-3, knowledge of a general algorithm that for any given $P=a\times G$ it can tell us for sure that $a$ is ODD/EVEN or $a$ is GREATER/SMALLER than $N/2$ or $a$ has $x$ bits will completely break the security of Elliptic Curves and by which it will be possible to retrieve $a$ from $P$.

But what about item 4? I mean if one can discover an algorithm by which they can determine that for any given $P$, $a$ is or is not a quadratic residue modulo $N$, will they be able to completely retrieve $a$ and break the cryptographic scheme?

What if the algorithm can also tell the square root of $a$ modulo $N$?

Update 1:

These questions arose when I was studying risks of private key database being partially compromised. That is what happens if an attacker knows clues of our private keys.

Answer 1

The impact on security from 1 and 2 can be considered as not significant. Say you have 256 Bit security, that implies $2^{256}$ different keys to chose from. If you know the key is odd, this is reduced to $2^{255}$ different keys. For 2 it is similar.

Knowing how many ones and zeros there are can be considered as a significant risk. Let's assume you have a 256 Bit key containing one 1. Only 256 possible keys remain. This results in the question of how many permutations there are. For binary numbers with a given length X and a given number if ones N this leads to the numbers of key, that can be calculated using:

$$ number\ of\ keys=\frac{(X!)}{ (N! * (X-N)!)} $$

In the best case this results in $\approx 2^{252}$ possible keys, for 128 ones. But for a different numbers of ones this only gets worse. Timing Attacks often try to find out how many different ones and zeros there are (or even try to locate some). These attacks are often a serious problem. Some of these attacks tried to guess the starting 1 of elliptic curve keys, because some old algorithms had different computing time depending on the position of the leading 1. This must not be a risk, e.g. the 1 is the first or second bit. But if the leading one was at position $i$ the security would drop to $2^{256-i}$. Depending on $i$ this can be a significant risk, especially if keys are completely chosen at random. You definitely don't want these risks.

For 4 I'm unsure. I will edit this, if I find a good answer.