Как объединить 3 таблицы в Postgres sql?

Я решал пример SQL Дэнни Ма для Pizza Runner. В одном из разделов «Цены и рейтинги» я застрял в двух вопросах и не могу прийти к правильному решению. Ссылка на кейс: https://8weeksqlchallenge.com/case-study-2/

[cleaned_customer_orders table](https://i.sstatic.net/fEgXb.png)
[cleaned_runner_orders table](https://i.sstatic.net/sl6kd.png)
[extras table](https://i.sstatic.net/Ggm0B.png)

CREATE TABLE customer_orders (  
  "order_id" INTEGER,  
  "customer_id" INTEGER,  
  "pizza_id" INTEGER,    
  "exclusions" VARCHAR(4),  
  "extras" VARCHAR(4),  
  "order_time" TIMESTAMP  
);  

INSERT INTO customer_orders  
  ("order_id", "customer_id", "pizza_id", "exclusions", "extras", "order_time")  
VALUES  
  ('1', '101', '1', '', '', '2020-01-01 18:05:02'),  
  ('2', '101', '1', '', '', '2020-01-01 19:00:52'),  
  ('3', '102', '1', '', '', '2020-01-02 23:51:23'),  
  ('3', '102', '2', '', NULL, '2020-01-02 23:51:23'),  
  ('4', '103', '1', '4', '', '2020-01-04 13:23:46'),  
  ('4', '103', '1', '4', '', '2020-01-04 13:23:46'),  
  ('4', '103', '2', '4', '', '2020-01-04 13:23:46'),  
  ('5', '104', '1', 'null', '1', '2020-01-08 21:00:29'),  
  ('6', '101', '2', 'null', 'null', '2020-01-08 21:03:13'),  
  ('7', '105', '2', 'null', '1', '2020-01-08 21:20:29'),  
  ('8', '102', '1', 'null', 'null', '2020-01-09 23:54:33'),  
  ('9', '103', '1', '4', '1, 5', '2020-01-10 11:22:59'),  
  ('10', '104', '1', 'null', 'null', '2020-01-11 18:34:49'),  
  ('10', '104', '1', '2, 6', '1, 4', '2020-01-11 18:34:49');  


CREATE TABLE runner_orders (
  "order_id" INTEGER,
  "runner_id" INTEGER,
  "pickup_time" VARCHAR(19),
  "distance" VARCHAR(7),
  "duration" VARCHAR(10),
  "cancellation" VARCHAR(23)
);

INSERT INTO runner_orders
  ("order_id", "runner_id", "pickup_time", "distance", "duration", "cancellation")
VALUES
  ('1', '1', '2020-01-01 18:15:34', '20km', '32 minutes', ''),
  ('2', '1', '2020-01-01 19:10:54', '20km', '27 minutes', ''),
  ('3', '1', '2020-01-03 00:12:37', '13.4km', '20 mins', NULL),
  ('4', '2', '2020-01-04 13:53:03', '23.4', '40', NULL),
  ('5', '3', '2020-01-08 21:10:57', '10', '15', NULL),
  ('6', '3', 'null', 'null', 'null', 'Restaurant Cancellation'),
  ('7', '2', '2020-01-08 21:30:45', '25km', '25mins', 'null'),
  ('8', '2', '2020-01-10 00:15:02', '23.4 km', '15 minute', 'null'),
  ('9', '2', 'null', 'null', 'null', 'Customer Cancellation'),
  ('10', '1', '2020-01-11 18:50:20', '10km', '10minutes', 'null');

select order_id, customer_id, pizza_id,
    case
        when exclusions in ('null', '') then null
        else exclusions
    end as exclusions,
    case 
        when extras in ('null', '') then null
        else extras
    end as extras,
    order_time
into cleaned_customer_orders
from customer_orders

select 
    order_id,
    runner_id,
    cast(case 
        when pickup_time = 'null' then null
        else pickup_time
    end as timestamp) as pickup_time,
    cast(case 
        when distance = 'null' then null
        else trim('km' from distance)
    end as float) as distance,
    cast(case
        when duration = 'null' then null
        else substring(duration, 1, 2)
    end as int)as duration,
    case
        when cancellation in ('null', '') then null
        else cancellation
end as cancellation
into cleaned_runner_orders
from runner_orders;

select * from cleaned_customer_orders
select * from cleaned_runner_orders

alter table cleaned_customer_orders
add column record_id serial;

select * from cleaned_customer_orders;

-- to generate extras table
select      
    cco.record_id,
    trim(e.value) as topping_id
into extras
from 
    cleaned_customer_orders as cco
    cross join lateral unnest(string_to_array(cco.extras, ',')) as e(value);

Вопрос 1. Запрос

select   
    sum(case when cco.pizza_id = 1 then 12 else 10 end) as pizza_cost  
from cleaned_customer_orders cco  
join pizza_names pn on pn.pizza_id = cco.pizza_id  
join cleaned_runner_orders cro on cro.order_id = cco.order_id  
where cro.cancellation is null  

В2. Запрос

with cte_extras_cost as (select  
    sum(case when pn.pizza_id = 1 then 12 else 10 end) as pizza_cost,  
    count (e.topping_id)*1 as extras_cost  
    from cleaned_customer_orders cco  
    join pizza_names pn on pn.pizza_id = cco.pizza_id  
    left join extras e on e.record_id = cco.record_id  
    join cleaned_runner_orders cro on cro.order_id = cco.order_id  
    where cro.cancellation is null)  
select pizza_cost, coalesce(extras_cost, 0) as extras_cost, (pizza_cost+coalesce(extras_cost, 0))  
 as total_cost  
from cte_extras_cost;  

Вопрос 5. Запрос

with cte as (  
    select   
    sum(cro.distance*0.30) as delivery_cost,  
    sum(case when cco.pizza_id = 1 then 12 else 10 end) as total_cost  
from cleaned_runner_orders cro  
join cleaned_customer_orders cco on cro.order_id = cco.order_id  
where cro.cancellation is null)  
select cte.total_cost, cte.delivery_cost, (cte.total_cost - cte.delivery_cost) as actual_cost  
from cte  

Мой результат:
Вопрос 1. 138
В2. Pizza_cost=150, extras_cost=4, total_cost=154
Вопрос 5. общая_стоимость=138, стоимость_доставки=64, фактическая_стоимость=73

Правильные ответы:
Вопрос 1. 138
В2. Pizza_cost=138, extras_cost=4, total_cost=142
Вопрос 5. общая_стоимость=138, стоимость_доставки=43, фактическая_стоимость=94