First lets start with A
What grade class is this?
In highschool, or for a quick estimate I would accept the (Ka * x)1/2 method.
But for real work I would expect people to use then
Ka = [H+] * [A-]
[HA]
*note* If you have not covered this before because you are in highschool, you can ignore it, but read it to see the % difference.
In your case that becomes:
( 7.224*10-4)2
0.04 - 7.224x10-4
= 1.336*10-5
Which is only 1.5% difference from what you got (before -log), but is still a better and more precise method of calculating.
And then B)
I am not sure what you are doing? This is a strong base.
You should know that:
Kw = [H+][OH-] = 1.0x10-14
You know Kw, and OH- because this is a strong base; one of the assumptions we are allowed to make is that a strong base (or acid) dissociates completely.
[H+] = Kw = 1.0x10-14
[OH-] 0.01
If you are in high school, please repost c and d in the high school forum, as in this forum people are expecting you to be in a college level analytical class. If you need more explanation for a and b just ask, but because you did not show what you tried for c and d I feel no desire to check if you are correct. It is more important to understand what/why then to just get the correct answer. Btw, question C you gave one answer, but from your question it should have two answers.