Needs["Graphics`Graphics`"]

Problem Number 1
r[t_] = (3  Cos[2  t])

PolarPlot[r[t], {t, 0, 2  Pi}]

Problem Number 2

PolarPlot[2 + Cos[4  t], {t, 0, 2  π}]

Part A

NIntegrate[
      2 + Cos[4  t], {t, 0, 2  π}]

Part B

S[t_] = 2 + Cos[4  t]

NIntegrate[
      Sqrt[(S[t])^2 + (S'[t])^2], {t, 0, 2  π}]

Problem number 3

PolarPlot[(1 + Cos[t])^2, {t, 0, 2  π}]

NIntegrate[(1 + Cos[t])^2, {t, 0, 2  π}]

Question Number 4

Part A

ListPlot[Table[n^10/(1.1)^n, {n, 0, 50}]]

It appears it might converge at y=0 because the graph slopes down at zero and seems to be increasing as x increases.

Part B

ListPlot[Table[n^10/(1.1)^n, {n, 0, 100}]]

Looking at this graph it seems that maybe this sequence will converge at y=7 x 10^15 or at least around there because the graph is begining to level of and may not increase any further.

Part C

ListPlot[Table[n^10/(1.1)^n, {n, 0, 200}]]

Looking at this graph I would have to go back to my original conclusion and state that the graph seems to be converging at zero due to the downward slopes towards zero.

Part D

F[n_] = n^10

G[n_] = 1.1^n

D[F[n], {n, 10}]

D[G[n], {n, 10}]

Limit[3628800/(6.18576*10^-11*1.1^n), 
        n -> ∞]

Question Number 5
Part A

ListPlot[Table[(4*5^n - 5*4^n)/6^n, {n, 1, 20}]]

ListPlot[
      Table[NSum[(4*5^n - 5*4^n)/6^n, {n, 1, i}], {i, 1, 20}]]

NSum[(4*5^n - 5*4^n)/6^n, {n, 1, 20}]

(∑_(n = 1)^∞)(4*5^n - 5*4^n)/6^n