Did you carefully read the mmap(2) man page? I recommend reading it several times.
Notice that you can only ask the kernel [thru mmap etc...] to manage memory aligned to and multiple of the page size sysconf(_SC_PAGE_SIZE)  which is often 4096 bytes (and I am supposing that in my answer).
Then you might do:
  size_t page_size =  sysconf(_SC_PAGE_SIZE);
  assert (page_size == 4096); // otherwise this code is wrong
  // 1000 bytes fit into 1*4096
  char *x = mmap (NULL, page_size, PROT_READ|PROT_WRITE, 
                  MAP_ANONYMOUS, -1, (off_t)0);
  if (x == MMAP_FAILED) perror("mmap x"), exit (EXIT_FAILURE);
  // 2000 bytes fit into 1*4096
  char *y = mmap (NULL, page_size, PROT_READ|PROT_WRITE, 
                  MAP_ANONYMOUS, -1, (off_t)0);
  if (y == MMAP_FAILED) perror("mmap y"), exit (EXIT_FAILURE);
later to free the memory, use
  if (munmap(x, page_size))
     perror("munmap x"), exit(EXIT_FAILURE);
etc
If you want to allocate 5Kbytes, you'll need two pages (because 5Kbytes < 2*4096 and 5Kbytes > 1*4096) i.e. mmap(NULL, 2*page_size, ...
Actually, all  of your x, y, z takes only 8000 bytes and could fit into two, not three, pages... But then you could only munmap that memory together.
Be aware that mmap is a system call which might be quite expensive. malloc implementations take care to avoid calling it too often, that is why they manage previously free-d zones to reuse them later (in further malloc-s) without any syscall. In practice, most malloc implementations manage differently big allocations (e.g. more than a megabyte), which are often mmap-ed at malloc and munmap-ed at free time.... You could study the source code of some malloc. The one from MUSL Libc might be easier to read than the Glibc malloc.
BTW, the file /proc/1234/maps is showing you the memory map of process of pid 1234. Try also cat /proc/self/maps in a terminal, it shows the memory map of that cat process.