正如注解中提到的，不允许向std名称空间添加任何附加符号，只有在显式允许时才专门化一些类。你的程序是病态的，但这不是它无法编译的原因。
在：

typename T = typename std::iterator_traits<Container::iterator>::value_type

::iterator依赖于Container模板参数，因此除非消除歧义，否则它将被解释为静态数据成员。那么替换将失败，T无法推导，并且您无法调用该函数。要解决这个问题，请使用typename消歧器：

typename T = typename std::iterator_traits<typename Container::iterator>::value_type

如果你经常编写这样的代码，定义一个类似于std::ranges::range_value_t的方便别名也是有意义的：

// note 1: we are not guaranteed to have a T::iterator alias, e.g. when T is an array
template <typename T>
using range_iter_t = decltype(std::begin(std::declval<T>()));

template <typename T>
using range_value_t = typename std::iterator_traits<range_iter_t<T>>::value_type;

// used like:

// note 2: conventional name is 'Range', not 'Container'
//         not every range is a container
template <class Range, typename Predicate, typename T = range_value_t<Range>>
// note 3: use perfect forwarding for r and and predicate because of one-time use
//         in the algorithm. std::ranges::find_if also takes Range&&
T find_first_or_default(Range&& r, Predicate&& predicate, T defaultVal = T()) 
{
    // note 4: avoid calling end() twice, end() isn't always trivial    
    // note 5: use std::begin/std::end to make this function work with C-style arrays
    auto end = std::end(r);
    // note 6: perfectly forward the predicate, don't just copy
    auto it = std::find_if(std::begin(r), end, std::forward<Predicate>(predicate));
    if (it == end)
        return defaultVal;
    return *it;
}
// note 7: it would also make sense to have a second overload which takes two
//         iterators, not just a function that takes a range