Это - ожидаемое поведение C# 4.0 равенства Кортежа?

For Reference Type: == performs an identity comparison, i.e. it will only return true if both references point to the same object. While Equals() method is expected to perform a value comparison, i.e. it will return true if the references point to objects that are equivalent.

For reference types where == has NOT been overloaded, it compares whether two references refer to the same object

By default, the operator == tests for reference equality, so Yes the result you are seeing is expected.

See Guidelines for Overriding Equals() and Operator == (C# Programming Guide):

In C#, there are two different kinds of equality: reference equality (also known as identity) and value equality. Value equality is the generally understood meaning of equality: it means that two objects contain the same values. For example, two integers with the value of 2 have value equality. Reference equality means that there are not two objects to compare.

By default, == (on a class) means reference equality; i.e. are they the same instance; what object.ReferenceEquals(x,y) would return.

You can provide your own == / != operators to get the expected behaviour - and when you override Equals it is important to override GetHashCode too (otherwise you break usage as a key - Why is it important to override GetHashCode when Equals method is overriden in C#?):

public static bool operator == (NameAndNumber x, NameAndNumber y) {
    if (x == null && y == null) return true;
    if (x == null || y == null) return false;
    return x.Number == y.Number && x.Name == y.Name;
    // or if polymorphism is important: return x.Equals(y);
}
public static bool operator !=(NameAndNumber x, NameAndNumber y) {
    return !(x == y); // lazy but works
}
public override int GetHashCode() {
    return (Name == null ? 0 : Name.GetHashCode()) +
        17 * Number.GetHashCode();
}