mysql 5.7 json filter and features products
For the first problem, AFAICT, there is no way to use SQL to project certain properties of a JSON object.  I think you will have to do that in the application layer.

For the second problem, how to search for products with a certain feature: I assume you have a table products with one row for each product.  The table contains a column of type JSON which contains the features of the product as you have shown.  If so, I think the following query should find the products where the value of the feature called "display" in English has the value of 5.5":

SELECT * FROM products 
WHERE JSON_EXTRACT(feature, 
    REPLACE(JSON_UNQUOTE(
        JSON_SEARCH(feature, 'one', 'display', NULL, '$.feature[*].en')),
        '.en', '.value')) = "5.5\"";


JSON_SEARCH will give you the path to the feature called 'display'.  To find the value of the same feature, we replace '.en' in the path with '.value' and use JSON_EXTRACT to look up the value at this path.  (JSON_UNQUOTE is needed to convert the JSON string from JSON_SEARCH to an SQL string to be passed to JSON_EXTRACT).