How to collect all js files without combining into one?

Question

gulp.js

Good afternoon, I assembled a assembler for myself on a video lesson on YouTube, and actually that's what interests me.
There is a single app.js file, which, when built, gets into the build folder with the name app.min.js, in order to connect jquery, I need to install it via npm and import it already in app.js, but I would like to just create it next to app.js is another file, called jquery.js, and include it already in the body as it is done with app.js, and, accordingly, with other js files (slick, etc.).
And another question, is it possible somehow during the assembly so that the file is not minified, and I still don’t understand why there is an “extra” webpack code in the app.min.js file? Screenshots below

Here is the assembly code:
js.js file in tasks

import webpack from "webpack-stream";

export const js = () => {
    return app.gulp.src(app.path.src.js, { sourcemaps: true })
        .pipe(app.plugins.plumber(
            app.plugins.notify.onError({
                title: "JS",
                message: "Error <%= error.message =>"
            })
        ))
        .pipe(webpack({
            mode: 'development',
            output: {
                filename: 'app.min.js',
            }
        }))
        .pipe(app.gulp.dest(app.path.build.js))
        .pipe(app.plugins.browsersync.stream());
}

PS Thanks in advance for your help, if you need any other files of the collector, then let me know