First of all, in the sample code, F8 is a mask for the carry, digit carry,
and zero bits.  This sample will clear those 3 bits only.

To access page 1 file registers, simply use this:
        bsf     STATUS,RP0
        mov**   file

To switch back to page 0 file registers, just clear the bit.  It's that
simple!!!
for the "h" (hex), I think it's just expressing the same number in 2
different formats.
Example: h'A4' = 0xA4.

As for the crystal, there's an internal /4 from crystal speed to actual
processor speed

----------
> From: Fernando Scalini <FScalini@AOL.COM>
> To: PICLIST@MITVMA.MIT.EDU
> Subject: memory terminology
> Date: Wednesday, November 26, 1997 12:19 AM
>
> Excuse my ignorance of such basic knowledge, but 3 books and several
weeks of
> banging my head against my desk have driven me to this.
>
> I can't figure out how memory is mapped for the PICs (16c84).  the data
> sheets show bank 0 from 00h to 7Fh and bank 1 from 80h to FFh.  I think
I've
> finally figured out the concept of data banks and register shadowing but
> their is a sample program in Predko's book that clears the STATUS flags
as
> follows:
>
> movlw 0x0F8
> andwf STATUS
>
> What does the "h" in the memory maps mean, and where does the final 8 in
> 0x0F8 fit in to this format.  Also, 0x0F8 was never defined previously in
the
> program.  why was this address selected and how can you know what its
value
> is.
>
> Another quick one:  If a processor is rated at 4Mhz, why use only a 1Mhz
> crystal.
>
> thanks for your help.