1. Comparing two different methods

Two different means of assessing the students have been suggested. First using a normal distribution and secondly using fixed percentages for each rating.

Using the normal distribution
A - > 2a
B - > a and < 2a
C - > -a and < a
D - > -2a and < -a
E - < -2a

Using the fixed percentage
A - > 80%
B - > 60% and < 80%
C - > 40% and < 60%
D - > 20% and < 40%
E - < 20%

The task is to compare and contrast the two different assessment procedures. You should consider the outcome if a teacher sets an easier test or a harder test.
You many assume that the standard deviation remains the same for each of the tests.

I have no idea where to start. Any help would be greatly appreciated.

NOTE: I have used "a" because i could not find sigma.

2. You should find the probability of each of those intervals using the normal distribution.

Like $P(Z>2)$

$P(1

$P(-1

$P(-2

$P(Z<-2)=P(Z>2)$

3. $P(Z>2) = 0.4772$ therefore $P(Z<-2) = 0.4772$
$P(12) - P(Z>1) = 0.4772 -0.3413 = 0.1359$ therefore $P(-2
$P(-11) = 0.3413 + 0.3413 = 0.6826$

4. P(Z>2) is not close to .5, it's about .02
without looking at the tables I'm sure it's .5-.4772.
Similarly, your third line is off too.
Your second line is wrong too.
P(Z>1) is larger than P(Z>2), which means you have a negative probabilty.
Think about what you wrote and look at the intervals.

5. P(Z>2) = 0.5 - 0.4772 = 0.0228
P(Z>1) = 0.5 - 0.3413 = 0.1587
P(1<Z<2) = P(Z>1) - P(Z>2) = 0.1359
P(-1<Z<1) = P(Z>-1) - P(Z>1) = (0.5 + 0.3413) - (0.5 - 0.3413) = 0.6826

This means P(Z>2) = 2.2% A
P(1<Z<2) = 13.6% B
P(-1<Z<1) = 68.3% C
P(-2<Z<-1) = 13.6% D
P(Z<-2) = 2.2% E

meaning majority of students would receive a "C" using normal distribution