2. (10 points) Five pieces of paper are written with $1$, $2$, $3$, $4$, and $5$, facing up from smallest to largest, stacked in a pile. Now, the 1, 3, and 5 are flipped to their backs, and still placed in their original positions. If the entire stack of paper is split at any one piece of paper into two stacks, and the upper stack is reversed and placed back on the lower stack, or if the entire stack of 5 pieces of paper is reversed, it counts as one "reversal". To transform the above arrangement of five pieces of paper so that they are all facing up, the minimum number of "reversals" required is $\qquad$. (The side with the number is the front)
2. (10 points) Five pieces of paper are written with $1, 2, 3, 4, 5$ respectively, facing upwards from smallest to largest, stacked into one pile. Now, the $1, 3,$ and $5$ are flipped to their backs and placed back in their original positions. If the entire stack is split at any one piece of paper into two stacks, and the upper stack is reversed and placed back on the lower stack, or if the entire stack of 5 pieces of paper is reversed, it counts as one "reversal". To transform the five pieces of paper from the above arrangement to all facing upwards, at least 5 "reversals" are required. (The side with the number is the front side)
【Analysis】First, consider the entire stack of paper as composed of 5 "sub-stacks", which can help determine the change in the number of "sub-stacks" after one reversal, leading to the conclusion.
【Solution】Solution: If we consider the entire stack of paper as being composed of parts that are all in the same direction: "sub-stacks", then the original entire stack of paper consists of 5 sub-stacks.
One "reversal" can only start from one piece of paper, placing the top piece of the upper part next to the top piece of the lower part, with the order in between remaining unchanged. After one reversal, the number of "sub-stacks" can only change in three ways: increase by 1, remain unchanged, or decrease by 1. When the entire stack is reversed, the number of "sub-stacks" does not change.

According to the requirement, to turn the 5 into the front side, there must be one reversal of all 5 pieces of paper, and the sequence of the entire stack remains unchanged. To turn 5 sub-stacks into one sub-stack, at least 4 reversals are required. Therefore, at least 5 reversals are needed.

The following explains that 5 reversals can ensure that all 5 pieces of paper are facing upwards: First, reverse the 1, then reverse 1 and 2 together, then reverse $2, 1, 3$, then reverse $3, 1, 2, 4$, and finally reverse $4, 2, 1, 3, 5$ to make them all face upwards, totaling 5 reversals (this is one of the possible sequences of reversals). Therefore, the answer is: 5.