OK to post! Thanks for the extension Dr. Z. 

Question 1:

GFseq(1/(1-x-x^2-x^3),x,30);
[1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080]


GFseq(1/(1-x+x^2-x^3),x,30);
[1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0]

GFseq(exp(-x)/(1-x),x,30);
[1, 0, 1/2, 1/3, 3/8, 11/30, 53/144, 103/280, 2119/5760, 16687/45360, 16481/44800, 1468457/3991680, 16019531/43545600, 63633137/172972800, 2467007773/6706022400, 34361893981/93405312000, 15549624751/42268262400, 8178130767479/22230464256000, 138547156531409/376610217984000, 92079694567171/250298560512000, 4282366656425369/11640679464960000, 72289643288657479/196503623737344000, 6563440628747948887/17841281393295360000, 39299278806015611311/106826515449937920000, 9923922230666898717143/26976017466662584320000, 79253545592131482810517/215433472824041472000000, 5934505493938805432851513/16131658445064225423360000, 14006262966463963871240459/38072970106357874688000000, 461572649528573755888451011/1254684545727217532928000000, 116167945043852116348068366947/315777214062132212662272000000, 3364864615063302680426807870189/9146650338351415815045120000000]

Question 2:
GFseq(x/(-x^2 - x + 1), x, 30);
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 

  987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 

  121393, 196418, 317811, 514229, 832040]
SeqRec([[1, 1], [0, 1]], 30);
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 

  987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 

  121393, 196418, 317811, 514229, 832040]

verify(%, `%%`);

 true

I used maple to check it. They are identical. 

The reason they are identical are because they both seem to have the same generating function. 

Question 3:

Define a(n) In terms of the generating function
Sum(a(n)*x^n, n=0..infinity) = 1/(1-3*x+5*x^2)

In other words a(n) is the coefficient of x^n in the Taylor expansion (around x=0) of 1/(1-3*x+5*x^2)

[In Maple, you can do coeff(taylor(1/(1-3*x+5*x^2),x=0,n+1),x,n) to compute it.]

Prove that the sequence a(n) satisfies the linear recurrence

a(n)=3*a(n-1)-5*a(n-2)


Answer 3:

I used maple to verify it using a couple of procedure albeit not valid proof and it does stand. 

f(x)=1/(1-3*x+5*x^2)
f(x) * (1-3x+4x^2) = 1
Now f(x) = a(0)+a(1)*x+a(2)*x^2..
....

Will have to look at the solutions, that violates the agreement of not peeking at the solutions.