We use casework on the first place student. It is impossible for $ A$ to be first, as this would agree with Prediction 1. If $ B$ is first, then $ (CB)$ could not have finished consecutively. So the possible pairs from Prediction 2 are $ (DA)$, $ (AE)$, and $ (EC)$. Only $ (DA)$ and $ (EC)$ are disjoint. We cannot start with $ BEC$, as this would place $ C$ as predicted in Prediction 1. So our order is $ BDAEC$, which has no agreement with Prediction 2. So $ B$ cannot be first. If $ C$ is first, then the possible pairs from Prediction 2 are $ (DA)$, $ (AE)$, and $ (CB)$. We cannot use $ (CB)$ because it would place $ B$ as predicted by Prediction 1. This leaves only $ (DA)$ and $ (AE)$, which are not disjoint, so it is impossible for $ C$ to be first. If $ D$ is first, we can either use $ (DA)$ or not. If we do, then we must choose either $ (EC)$ or $ (CB)$ as our second pair. If we use $ (EC)$, we cannot start $ DAEC$, as this agrees in four places with Prediction 2. So we have $ DABEC$, which shares $ (AB)$ with Prediction 1, so it is impossible. If we use $ (CB)$, we cannot start $ DACB$, as $ C$ agrees with Prediction 1. So we are left with $ DAECB$, which agrees in five places with Prediction 2. So we cannot use $ (DA)$. The possible pairs from Prediction 2 are $ (AE)$, $ (EC)$, and $ (CB)$. Only $ (AE)$ and $ (CB)$ are disjoint. We cannot place $ (AE)$ right after $ D$, as $ (DA)$ is forbidden. So we must have $ DCBAE$, in which $ E$ agrees with Prediction 1. So $ D$ cannot be first. Thus, $ E$ is first. The possible pairs from Prediction 2 are $ (DA)$, $ (EC)$, and $ (CB)$. If we use $ (EC)$, then we must also use $ (DA)$. But $ ECBDA$ and $ ECDAB$ have only zero and one correspondences with Prediction 2, respectively. So we cannot use $ (EC)$ If we don't use $ (EC)$, then we must have $ EDACB$ (preventing $ (EC)$ by not placing $ (CB)$ right after $ E$). This clearly satisfies all requirements, so it is the desired order.