Answer:
a) We can find the z score for the value of 225 and we got:

And we can find this probability with the complement rule:

b) 
And we can find this probability using the normal standard table or excel and we got:

c) 
And we can find this probability with this difference:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

d) For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.8 of the area on the left and 0.2 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the costs of a population, and for this case we know the distribution for X is given by:
Where
and 
We are interested on this probability

And the best way to solve this problem is using the normal standard distribution and the z score given by:

We can find the z score for the value of 225 and we got:

And we can find this probability with the complement rule:

Part b

And the best way to solve this problem is using the normal standard distribution and the z score given by:

If we apply this formula to our probability we got this:

And we can find this probability using the normal standard table or excel and we got:

Part c

And the best way to solve this problem is using the normal standard distribution and the z score given by:

If we apply this formula to our probability we got this:

And we can find this probability with this difference:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

Part d
For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.8 of the area on the left and 0.2 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got