How to solve this probability problem using events Here is the problem with the work I've done with each question.
There is work to do on a building. The jobs will be done by guy A or guy B.     
The probability that A is available is 0.65.     
The probability that B will be available is 0.70.     
The probability that none of the two guys will be available is 0.2.



*

*Let A be the event "guy A is available"  

*Let B be the event "guy B is available"

*Let N be the event "None available"


A) Find the probability that both of the guys will be available. Show it with events.
$P(A|\bar{N}) * P(B|\bar{N}) = 0.65*0.8 * 0.7*0.8 = 0.29 $
My reasoning behind this is : I want the opposite of N, multiplied by the prop. of A and B, and then I put those together.
B) Find the probability that at least one guy will be available.
$1 - P(\bar{A}|N) * P(\bar{B}|N) = 0.35*0.2 + 0.3*0.2 = 0.13$
$ 1-0.13 = 0.87  $
My reasoning behind this is : 1 - the probability that none are available.
C) Find the probability that one and only one will be available.
$P(A|\bar{N}) * P(\bar{B}|\bar{N}) + P(\bar{A}|\bar{N}) * P(B|\bar{N})$
$0.65*0.8 * 0.3*0.8 + 0.35*0.8 * 0.7*0.8 = 0.28$
Thoughts : probab of A avail. and B not + probab of B avai. and A not.
D) Find the probability that guy B will be available, knowing that A is not available.
Well... I could use some help to do this one.
 A: I would draw a Venn diagram

B) Find the probability that at least one guy will be available.
You just want
$$P(A \cup B) = 1 - P(N)$$
I agree with your reasoning "1 $-$ the probability that none are available" here but not your answer.
A) Find the probability that both of the guys will be available. Show it with events.
You want $P(A\cap B)$, which I have written $P(AB)$ on my diagram, given by
$$P(A \cup B) = P(A) + P(B) - P(A\cap B)$$
rearranging this equation gives
$$P(A\cap B) = P(A) + P(B) - P(A \cup B)$$
you can also substitute $P(A \cup B) = 1 - P(N)$ from part B, above, to express $P(A\cap B)$ directly in terms of the probabilities you are given in the question
$$P(A\cap B) = P(A) + P(B) + P(N) - 1$$
C) Find the probability that one and only one will be available.
You want A or B but not both
$$P(A \cup B) - P(AB)$$
D) Find the probability that guy B will be available, knowing that A is not available.
This is conditional probability, so you can't just rely on the diagram: you need the formula.
$$P(B|\neg A) = \frac{P(B \cap \neg A)}{P(\neg A)}$$
You can use the diagram to get the numerator and the denominator.
$$P(B \cap \neg A) = P(B) - P(AB)$$
and
$$P(\neg A) = 1 - P(A)$$
Commentary on conditional probability
David's answer makes some great points about conditional probability and dependence of events in general, which I will not rehearse again here. Suffice to say that I agree that a simple approach is possible for the first three questions.
A: You are using conditional probability notation for your answers.  I'm not sure if that's what you actually meant, as none of the three answers involves conditional probability at all.
Conditional probability is used in the situation where you already have some information about what has happened, and you want to calculate a probability which takes that into account.  The word normally used is "given", as in "find the probability that A occurs, given that B has occurred".  Other terminology can be used, and in fact "knowing" in your fourth question serves the same purpose: it could be rephrased as "find the probability that B is available, given that A is not available".  So (D) is in fact a conditional probability question.
Actually (B) is the easiest question: your reason is correct, but where did the number $0.13$ come from? - the probability that none is available is given and you don't need to calculate it.
The simplest way to visualise (A), (B) and (C) is to draw a Venn diagram.  You should also know the "inclusion/exclusion" formula
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)\ .$$
This together with your answer to (B) will give $P(A\cap B)$, which will be the answer to (A).  For (C) your idea is correct but the numbers are wrong.
Note that you cannot always multiply probabilities: the formula
$$P(A\cap B)=P(A)P(B)$$
is true only if $A$ and $B$ are independent.  You should never assume two events are independent unless you have a specific reason for it.  In this case it is very easy to imagine that the availabilities of A and B are not independent: if there is a lot of building work going around then probably both will be unavailable, while if there is not much then probably both will be available.
The conditional probability formula is
$$P(X|Y)=\frac{P(X\cap Y)}{P(Y)}$$
and once you identify what you mean by $X$ and $Y$, this is what you need for (D).  Good luck!
