Windows 7 Ultimate

The fact that it shows "Installed memory (RAM): 4.0 GB (3.5 GB usable)" tells me you're running a 32-bit kernel.

Not necessarily. Depending on your motherboard, that can still happen even with a 64 bit OS.

There's an easier way (which doesn't depend on having the 64 bit C++ compiler installed, which iirc VS doesn't install by default even on x64):

using System;

namespace PointerSizeTest
{
    static class Program
    {
        static void Main()
        {
            Console.WriteLine("You are running a {0} bit operating system.", IntPtr.Size * 8);
        }
    }
}

.Net applications compiled with CPU target AnyCPU (which is the default) will run as 32 bit on a 32 bit OS, and as 64 bit on a 64 bit OS. Therefore, this will give you the correct answer.

In C++ I'd do this:

#define WIN32_MEAN_AND_LEAN
#include <windows.h>
#include <iostream>
using namespace std;

int main()
{
    SYSTEM_INFO info;
    GetNativeSystemInfo(&info);

    cout << "You are running Windows with a ";
    switch( info.wProcessorArchitecture )
    {
    case PROCESSOR_ARCHITECTURE_INTEL:
        cout << "x86";
        break;
    case PROCESSOR_ARCHITECTURE_AMD64:
        cout << "x64";
        break;
    case PROCESSOR_ARCHITECTURE_IA64:
        cout << "Itanium";
        break;
    default:
        cout << "unknown";
        break;
    }

    cout << " processor architecture." << endl;

    return 0;
}

Which, because it uses GetNativeSystemInfo, will give you the correct answer even if you compile it as a 32 bit executable.

Will that report x86 on a 64-bit processor running 32-bit Windows?

Yes, it reports the architecture used by Windows, not that of the CPU itself (if they are different).

No, it shouldn't fail. I think you get a few hardware changes for free before Internet activation starts failing, and even then only if you do it too frequently I believe. But even if it fails, you just call them, tell them you're transferring the license to another PC, and they should activate it for you. Takes five minutes at most.