Using pathlib, if you have a {python}base_fp, how would you create a {python}ressource_fp (just the filepath, not creating a file or folder), that is simply the path to the folder {python}"ressources" at {python}base_fp?
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from pathlib import Path

base_fp = Path("...")
ressource_fp = base_fp / "ressources"

# another option
base_fp.joinpath("ressources")

Using pathlib, please get the current working directory (the directory, where your python script is located).
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from patlib import Path

print(Path.cwd())

Using pathlib, please iterate through all files and folders at {python}filepath. Call {python}handle_files(file)if it is a file and {python}handle_folder(folder)if it is a folder.
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from pathlib import Path
filepath = Path('/path/to/your/folder')

# This loop will itearte through 
for item in filepath.iterdir():
    if item.is_file():
        handle_file(item)

	if item.is_dir():
		handle_folder(item)

Using pathlib, please {python}print("file with pattern found") for each file in folder {python}folder_fp, that fits regex pattern {python}r_pattern. Please ignore lower/higher case.
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from pathlib import Path
import re

for item in folder_fp.iterdir():
	if item.is_file():
		filename = item.stem
		match = re.search(pattern=pattern, string=filename, flags=re.IGNORECASE)
		if match:
			print("file with pattern found")