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]

2. 
Yes, they both return [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].
The generating function x/(1-x-x^2) satisfies the linear refurrence that is equivalent to the Fibonacci sequence. 

3. 
1, 3, 4, -3, -29, -72, -71, 147, 796, 1653, 979, -5328, -20879, -35997, -3596, 169197, 525571, 730728, -435671, ... , f^n(0)/n! where f = 1/(1-3*x+5*x^2)


Sum(a(n)*x^n, n=0..infinity) = 1/(1-3x + 5x^2)
f(x) = 1/(1-3x+5x^2)
(1 - 3x + 5x^2) * f(x) = 1 
f(x) = 1 + 3xf(x) - 5x^2f(x)
substitute f(x) = a(0) + a(1)x + a(2)x^2 + .... + a(n)x^n
f(x) = 1 + 3x*(a(0) + a(1)x + a(2)x^2 + ... + a(n)x^n) - 5x^2*(a(0) + a(1)x + a(2)x^2 + ... + a(n)x^n)
f(x) = 1 + 3x*a(0) + 3x^2a(1) + .... + 3x^(n+1)a(n) - 5x^2a(0) - 5x^3a(1) + ... + 5x^(n+2)a(n)
f(x) = 1 + 3xa(0) - 5x^2a(0) + 3x^2a(1) - 5x^3a(1) + 3x^3a(2)+ ....+ 3x^(n)a(n-1) - 5x^(n)a(n-2) + 3x^(n+1)a(n) - 5x^(n+1)a(n-1) + 5x^(n+2)a(n)
Looking at the leading coefficients of x^n, we can see that a(n) = 3a(n-1) - 5a(n-2).


Sum(a(n)*x^n, n=0..infinity) = 1/(1-c1x-... -ckx^k)
f(x) = 1/(1-c1x- ... -ckx^k)
(1-c1x-... -ckx^k)f(x) = 1
(1-c1x-... -ckx^k)(a(0) + a(1)x + a(2)x^2 + ... + a(n)x^n) = 1
a(0) + (a(1) - c1a(0))x + (a(2) - c1a(1) - c2a(0))x^2 + ... + (a(k) - c1a(k-1) - ... - cka(0))x^k + ... + (a(n) - c1a(n-1) - ... - cka(n-k))x^n = 1
(a(0) - 1) + (a(1) - c1a(0))x + (a(2) - c1a(1) - c2a(0))x^2 + ... + (a(k) - c1a(k-1) - ... - cka(0))x^k + ... + (a(n) - c1a(n-1) - ... - cka(n-k))x^n = 0
Therefore, a(n) = c1a(n-1) + c2a(n-2) + ... + cka(n-k).