Secret sharing scheme WITHOUT Shamir Secrete Sharing I am planning to use the following XOR scheme to divide a secret into only 2 shares (I do not want to use Shamir Secret Sharing for different reasons that are beyond the scope of this post).

Here's an example of the XOR scheme I have in mind.

SECRET (128 bits)  =  0 1 0 1 0 0 1 0 0 0 0 1  (…)  0 0 1 1 0

SHARE 1 =  TRUNC128 (SHA512 (SECRET))

     =  0 0 0 0 1 1 1 0 0 0 1 1  (…)  1 1 0 0 1


SHARE 2 =  SECRET ^ SHARE1

     =  0 1 0 1 0 0 1 0 0 0 0 1  (…)  0 0 1 1 0

     ^  0 0 0 0 1 1 1 0 0 0 1 1  (…)  1 1 0 0 1

    =  0 1 0 1 1 1 0 0 0 0 1 0  1 1 1 1 1


SECRET   =  SHARE1 ^ SHARE 2

   =  0 0 0 0 1 1 1 0 0 0 1 1  (…)  1 1 0 0 1

    ^  0 1 0 1 1 1 0 0 0 0 1 0  (…)  1 1 1 1 1

   =  0 1 0 1 0 0 1 0 0 0 0 1  (…)  0 0 1 1 0


I have a few questions about that XOR scheme as I am building a case for using it.

1) How resistant would such a scheme be to common attack vectors (brute force...)?

2) Are there any ways to evaluate what it (or how much it) would take to break that scheme?

3) Is that scheme already used in "notably safe applications"? If that is the case, which ones?
XOR secret sharing used directly is unconditionally secure, if the randomly generated first share $S_1$ is a uniformly distributed vector in $\{0,1\}^{128}$. Let the secret be $X \in \{0,1\}^{128}$ and the shares be $S_2=X+S_1,$ and $S_1,$ where $+$ denotes bitwise XOR.

For a third party to learn the secret, they need to brute force $O(2^{128})$ guesses in the worst case.

Now you've "hidden" the secret $X$ by your $S_1'=TRUNC_{128}(SHA_{512}(X)).$

As stated in the comments I was wrong, thebirthday paradox does not directly apply.

An attacker brute forcing the share $S_2’=X+X_1$ at the same complexity will not learn your secret but its hash, so you’ve gained something.

However in the absence of side information, the original XOR scheme is as secure, unless the guesses for the secret can be tested by using it, say, as a password to unlock some system. In that use case, your modification is an improvement since it prevents this way of testing if the guess is correct.