Does knowledge of SHA256(secret ∥ known constant) in addition to SHA256(secret) give any information about the secret? Does knowledge of SHA256(secret) and SHA256(secret ∥ known constant) give any more information about the secret than just knowledge of SHA256(secret)?
In the first case; the attacker has this knowledge


$\operatorname{SHA256}$
$x = \operatorname{SHA256}(\text{secret})$


While brute-forcing this hash value $x$, there is one problem. One cannot determine whether this is the pre-image or it is a second pre-image. For password cracking, that is not really important since you find a valid password for the target. For your case, assuming the secret has to be determined exactly, there is no way for the attacker. Actually, finding more than one break the collision property of SHA256, see in Fgrieu's answer. Therefore, it is hard to find more than one.

In the second case; the attacker has


$\operatorname{SHA256}$
$x = \operatorname{SHA256}(\text{secret})$
$\text{the known constant}$
$y = \operatorname{SHA256}(\text{secret}\mathbin\|\text{the known constant})$


This additional knowledge gives the attacker extra information that can be exploited. While brute-forcing to find the secret, this time the attacker has a new mechanism to distinguish the second pre-images from the secret. Given a secret candidate $candidate_s$ check;

$$ y = \operatorname{SHA256}(candidate_s\mathbin\|\text{the known constant})$$ 

Apart from the pre-image, the probability of this equality is negligible since 
 we expect that SHA-256 has collision resistance.

As a result, this gives extra information about the secret that can be exploited.

Side note: if the entropy of the $secret$ is larger than 128-bit, it is not possible to execute brute-force on-the secret.