Go 言語で分散タスクキューと同時タスクのタスクスケジューリング戦略の問題を解決するにはどうすればよいですか?

package main

import (
    "fmt"
    "github.com/go-redis/redis/v8"
    "time"
)

func main() {
    client := redis.NewClient(&redis.Options{
        Addr:     "localhost:6379",
        Password: "",
        DB:       0,
    })

    // 生产者将任务添加到队列中
    err := client.LPush(context.Background(), "task_queue", "task1", "task2").Err()
    if err != nil {
        fmt.Println(err)
        return
    }

    // 消费者从队列中获取任务并执行
    for {
        res, err := client.BRPop(context.Background(), 0, "task_queue").Result()
        if err != nil {
            fmt.Println(err)
            return
        }

        fmt.Println("Processing task:", res[1])
        time.Sleep(time.Second)
    }
}

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {
    go func() {
        fmt.Println("Task 1 started")
        time.Sleep(time.Second)
        fmt.Println("Task 1 finished")
    }()

    go func() {
        fmt.Println("Task 2 started")
        time.Sleep(time.Second)
        fmt.Println("Task 2 finished")
    }()

    // 等待所有任务完成
    var wg sync.WaitGroup
    wg.Add(2)
    wg.Wait()

    fmt.Println("All tasks completed")
}

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {
    var wg sync.WaitGroup

    for i := 0; i < 10; i++ {
        wg.Add(1)

        go func(taskNum int) {
            defer wg.Done()

            fmt.Printf("Task %d started
", taskNum)
            time.Sleep(time.Second)
            fmt.Printf("Task %d finished
", taskNum)
        }(i + 1)
    }

    wg.Wait()

    fmt.Println("All tasks completed")
}