You can’t simply multiply the probabilities of $E_1$ and $E_2$, because what happens in the even-numbered positions isn’t independent of what happens in the odd-numbered positions. For instance, it’s clearly possible to get matches in exactly four of the even-numbered positions, so $P(E_1)\ne 0$, but if you have matches in all five odd-numbered positions, there is absolutely no way for $E_1$ to occur: you can’t have exactly $9$ matches.
Your calculations of $P(E_1)$ and $P(E_2)$ are also incorrect. Let’s look at $E_1$. There are $\binom54=5$ ways to choose which four of the even-numbered positions will be matches. There are $5$ ways to choose which odd-numbered card will fill the other even-numbered position, and there are $5!$ ways to permute the remaining $5$ cards among the $5$ odd-numbered positions. Thus, there are $5\cdot 5\ \cdot 5! = 25\cdot 120 = 3000$ permutations of the deck that result in $E_1$, and therefore $P(E_1) = \frac{3000}{10!} = \frac{5}{6048},$ since there are $10!$ different ways to deal out the cards, all of them equally likely.
If $n$ is the number of ways to get matches in exactly four of the even-numbered positions and three of the odd-numbered positions, the desired probability is $\dfrac{n}{10!}$, so let’s calculate $n$.
There are $\binom54 = 5$ ways to choose which four of the five even-numbered positions will be matches, and there are $\binom53 = 10$ ways to choose which three of the five odd-numbered positions will be matches, so there are $5\cdot 10 = 50$ ways to specify the seven matching positions.
This leaves three positions, say $p_1,p_2$, and $p_3$, and the three cards numbered $p_1,p_2$, and $p_3$. These three cards will end up in these three positions, so we have to make sure that none of them ends up in the ‘right’ position. There are $3! = 6$ possible permutations of cards $p_1,p_2$, and $p_3$ in positions $p_1,p_2$, and $p_3$, and it’s easy to see that only the permutations $p_3,p_1,p_2$ and $p_2,p_3,p_1$ avoid a match. Thus, for each of the $50$ ways to specify the seven matching positions, there are $2$ ways to fill in the other three positions without getting another match. This means that there are just $50\cdot 2 = 100$ permutations of the deck that produce the desired set of matches: $n = 100$, and the desired probability is $\frac{100}{10!} = \frac{1}{36288}.$