Deduction of the function

Thanks to the adoption of template parameter deduction for constructors, in C++17, you'll be able to just write:

A a(::close, 1);

Before that, you'll just need to write a factory to do the deduction for you:

template <class F, class... Args>
A<std::decay_t<F>> make_a(F&& f, Args&&... args) {
    return {std::forward<F>(f), std::forward<Args>(args)...};
}

auto a = make_a(::close, 1);

This is a little verbose, but at least you don't need to worry about efficiency - there will be no copies made here thanks to RVO.

You cannot omit the arguments of a template class, unless they are defaulted. What you can do is have a maker function which deduces the argument and forwards this argument to the template class, returning an object of the appropriate instantiation.

template<typename F, typename... Args>
A<F> make_A(F f, Args&&... args) {
    return A<F>(f, std::forward<Args>(args)...);
}

No, you (currently) cannot. The standard way of doing this is by creating "make_like" function (such as make_pair, make_optional ...):

template<typename F, typename... Args>
A<std::decay_t<F>> make_A (F &&f, Args&&... args) {
    return {std::forward<F>(f), std::forward<Args>(args)...};
}

C++17 will introduce template argument deduction for class which will allow you to do exactly what you want (see also Barry's answer below).