Answer
$ a.\quad$
See image. Steps given below.
$ b.\quad$
$N=4$
$N=6$
Work Step by Step
$a_{1}=1$
$a_{2}=1+\displaystyle \frac{1}{5}$
$a_{3}=1+\displaystyle \frac{1}{5}+\frac{1}{25}$
$a_{4}=1+\displaystyle \frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}$
$...$
$a_{n}=\displaystyle \sum_{k=1}^{n}(\frac{1}{5})^{n-1}$ = sum of the first n terms of a geometric sequence
$a_{n}= \displaystyle \frac{1-(\frac{1}{5})^{k}}{1-\frac{1}{5}}=\frac{5(1-5^{-k})}{4}$
$a.\quad $
The steps you take will depend on the CAS you are using, but they follow the same logic.
Using the free online CAS at geogebra.org/cas:
Cell 1: Enter the function representing the sequence
$a(x)=\displaystyle \frac{5(1-5^{-k})}{4}$
From the dropdown menu, select "Table of values".
In the dialog box for the table, set the range from 1 to 25, step 1.
When we observe the graph, it seems to be increasing
towards a certain value $\approx 1.25$, (never above it.)
The sequence seems to converge.
In the next free cell of the CAS, we find the limit when $ n\rightarrow\infty$
Here, we enter "L=Limit(a, infinity)" (without quotes)
The CAS returns the limit to be $\displaystyle \frac{5}{4}=1.25.$
$b.\quad $
For $\epsilon=0.01$
We find the intersection of the graph of a(x) with the horizontal line
$y=L-0.01$
Enter: Solve(L-a(x)=0.01)
The CAS returns $x=3.$
We take the next integer, $N=4.$
We find the intersection of the graph of a(x) with the horizontal line
$y=L-0.0001$
Enter: Solve(L-a(x)=0.0001)
The CAS returns $x=5.86....$
We take the next integer, $N=6.$
