Mixed Strategy Nash Equilibrium for this particular 3x3 matrix
You generally have the right approach with expected payoffs correctly, but there are some errors in your logic.
First, take a look at your reduced matrix. You're missing a way to reduce it further and simplify the math a bit! You've left an extra strategy in place for Player 2. Player 2 knows that Player 1 will never play $M$, so Player 2 only needs to consider best responses to $L$ and $R$.
Second, note that you're using Player 1's payoffs in your expected payoff calculations for Player 2. For example, you have
$$E[A] = \frac{1}{2}q + 1(1-q)$$
The payoffs of $\frac{1}{2}$ and $1$ are to Player 1, not to Player 2. You want to use $\frac{1}{2}$ and $0$ in your calculation.
Third, the way you find a mixed strategy Nash equilibrium is by setting players' expected payoffs to be equal. Player 1 and Player 2 each need to be indifferent between their strategies, and that occurs when they play their strategies with probabilities so that the other player's payoffs are equal.
So, for Player 2's equilibrium strategy, you'll take Player 1's expected payoffs and set them equal: $E[L] = E[R]$. If they aren't equal, then one's greater than the other. Say $E[L] &gt; E[R]$. Then, Player 1 will play $L$, and Player 2 will want to play $C$, then Player 1 will want to play $R$, and you end up in a cycle (and not an equilibrium). When the payoffs are equal, you have a stable solution, and neither player has an incentive to deviate from that strategy.
The same rationale applies to Player 1's equilibrium strategy - it's the set of probabilities that result in all of Player 2's expected payoffs being equal, and your Nash equilibrium is the set of all the strategies and their corresponding probabilities.