Dear NumberTheory Scholar-Apprentices.  

Question Q1:
 `Compute the last digit of 37^{471} and its last TWO digits.'

Myself, I would rather write than compute, so I will organize my
thoughts so that I hardly need compute, by making a lot of bitsy steps.

Firstly, since B := 37 is coprime to 10 and to 100, the EFT tells us that
	P := Per_{10}(B)   <|  phi(10)  =note=  4,   and
	Q := Per_{100}(B)  <|  phi(100) =note= 40.

Also, ten divides one-hundred, so

1:	P <| Q <| 40.

  Now let == mean cong.mod.10.  Our P is either 1,2, or 4, but it is
neither 1 nor 2 since,  B^1 == -3 which is not ==1, and

	B^2 == [-3]^2 = 3^2 == -1,

which is not 1.   Thus the mod-10 period of B is 4.


  The mod-10 cycle that we see for B is

    n:	0       1          2      3      4
  B^n:	1  ->  -3  ->  9==-1  ->  3  ->  1

where "->" means multiplication by B, mod 10.


  Reducing the exponent, 471, modulo 4 yields 3, so

	B^{471} == B^3 == 3.

  
LAST TWO DIGITs:  From (1), the mod-100 period, Q, is either

	4, 8, 20 or 40.

NOW let == mean cong.mod.100.  I now compute mod-100, writing my
results to end in digit "1", so that I can do all the computations
trivially in my head [as opposed to out of my mind].  I put "NO" next
to each POTENTIAL value of Q, which turned out NOT to be Q.

	 n:	<B^n>_{100}	==	Std.Form
 -----------------------------------------------
	 1:	37 ==			 37
	 2:	69 ==			-31
  NO	 4:	[-31]^2 = [31]^2 ==	 61
  NO	 8:	[61]^2 ==		 21
	16:	[21]^2 ==		 41
 YES	20:	B^4 * B^{16} == 61*41 == 01 

Then 471 =20= 71 =20= 11.  (You guessed what "=20=" means, yes?)

  Now note that eleven equals 8+2+1.  So our table continues as
	11:	21*[-31]*37 == 21*31*[-37] == 51*[63] == 13 .

Consequently,
  LastTwoDigits(37^{471}) = "13" .

		--Best Wishes, "Prof. Jonathan"