Warning: mysqli_select_db() expects exactly 2 parameters, 1 given

mysqli_select_db() should have 2 parameters, the connection link and the database name -

mysqli_select_db($con, 'phpcadet') or die(mysqli_error($con));

Using mysqli_error in the die statement will tell you exactly what is wrong as opposed to a generic error message.

This error message is telling you that you need to provide the mysqli connection object as the first argument to the mysqli_select_db() function. Most of mysqli functions require the mysqli object when used in the procedural style.

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$con = mysqli_connect('localhost', 'root', 'PwdSQL5');
mysqli_select_db($con, 'phpcadet')
//               ^^^^ - pass the $con object from the line above

However, you don't need to use mysqli_select_db() at all. You can pass the database name as the fourth argument to the mysqli_connect() function. All you need is this:

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$con = mysqli_connect('localhost', 'root', 'PwdSQL5', 'phpcadet');
//                                  the database name - ^^^^^^