Find encryption/auth algorithm name - Experts!

I'm looking for an expert in cryptography to find the right encryption algorithm from an existing device. We're designing an electronic board to talk with another device. We have almost 90% of the communication done, but there is an initial authorization talk with the device where we cannot be sure the type of encryption or key-exchange algorithm. We're looking for a cryptography expert to tell us the algorithm used on that encryption/key-exchange, the keys found and a detailed procedure on how we can get those keys. The communication procedure is: - We have device A (the machine with the algorithm) and device B (our board in the end) - Device A reads data from device B such description, Id, etc - Device A sends 64 bytes (different each time it is initialized) - Device B sends back 256 bytes (different each time as they seems to be correlated to previous point) - Device A sends back 256 bytes again That's what we are looking forward to discover. We will be able to successfully test if you are succeed because Device A then use the exchanged key to send another command (later on) and Device B should answer back with the correct encrypted value. That's the end of the project. The programmer/engineer for this project should let us know how to get the keys telling us exactly what we should program / get / fool device A in order to get the keys and then use that for program our device B correctly as we know the full other communication messages. We have about of 100+ valid full authentications from the real device A/device B connection. We will provide them. We suspect it is a key-exchange algorithm such Diffie-Hellman or RSA. We know it can be fooled, so we are looking the expert to tell us how. We will pay in full once the awarded engineer provides us a working example based on our captures of the messages. Please write 'caramel' word in your description so we know you read it all correctly and understand. No copy/paste bids will be considered. We will pay up to 1000 USD. Thank you!