The optimization process and recommendations:

1. Avoid Correlated Subqueries In Select Clause (modified query below): The aggregation function located in a subquery inside the SELECT clause, is executed once for every matched row. Extracting this subquery to a temporary table will improve performance significantly.
2. Create Optimal Indexes (modified query below): The recommended indexes are an integral part of this optimization effort and should be created before testing the execution duration of the optimized query.
3. Explicitly ORDER BY After GROUP BY (modified query below): By default, the database sorts all 'GROUP BY col1, col2, ...' queries as if you specified 'ORDER BY col1, col2, ...' in the query as well. If a query includes a GROUP BY clause but you want to avoid the overhead of sorting the result, you can suppress sorting by specifying 'ORDER BY NULL'.

Optimal indexes for this query:

ALTER TABLE `es_temp1` ADD INDEX `es_temp1_idx_letter_to` (`letter_to`);
ALTER TABLE `letters` ADD INDEX `letters_idx_letter_from_letter_to` (`letter_from`,`letter_to`);
ALTER TABLE `letters` ADD INDEX `letters_idx_letter_status_letter_read_da` (`letter_from`,`status`,`letter_to`,`read_date`);
ALTER TABLE `letters` ADD INDEX `letters_idx_read_date` (`read_date`);

The optimized query:

SELECT
        let.letter_from,
        let.`letter_to`,
        COUNT(let.`letter_from`) AS sendMail,
        es_temp1.readMail,
        (SELECT
            letters.`read_date` 
        FROM
            letters 
        WHERE
            letters.letter_from = 310 
            AND letters.`status` = 1 
            AND letters.`letter_to` = let.`letter_to` 
        ORDER BY
            letters.`read_date` DESC LIMIT 1) AS dateLastRead,
        (SELECT
            `auto` 
        FROM
            letters 
        WHERE
            letters.letter_from = 310 
            AND letters.`status` = 1 
            AND letters.`letter_to` = let.`letter_to` 
        ORDER BY
            letters.`read_date` DESC LIMIT 1) AS auto 
    FROM
        `letters` let 
    LEFT JOIN
        es_temp1 
            ON es_temp1.letter_to = let.`letter_to` 
    WHERE
        let.letter_from = 310 
    GROUP BY
        let.letter_to 
    ORDER BY
        NULL