Answer
$ a.\quad$
See image. Steps given below.
$ b.\quad$
$N=21$
$N=2070$
Work Step by Step
$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
$a(x)=(1+0.5/x)^{\wedge}x$
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$1.65.
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 $\sqrt{e}\approx 1.64872127.$
$b.\quad $
For $\epsilon=0.01$
we find the intersection of the graph of a(x) with the horizontal line
$y=\sqrt{e}-0.01$
($y=1.63872127$)
but, the CAS encounters precision limits due to the number of of calculations.
So, we pursue another avenue.
The function rises. We observe the table of points that was created and note that
$a_{21}=1.639088747454687$,
which is first below the $0.01$ distance from the limit.
We take $N=21$.
For $\epsilon=0.0001$
we want to find the first term of the sequence above
$\sqrt{e}-0.0001\approx 1.648(6)2127$
Delete the list of values (remove column a(x) from the table)
and create a new one, ranging from 100 to 3000, step 10.
We find that
$a_{2060}=1.648621246153068,\quad |a_{2010}-L|=0.00010002454706$
$a_{2070}=1.648621729271136,\quad |a_{2010}-L|=0.000099541428992$
so we take $N=2070$
