Probability for not finding a product 
A product is exist in $\frac{1}{4}$ of chain stores. One decided looking for the product in not more than 5 stores. Defining $X$ to be number of searched stores, find:
a.Distribution of $X$
b.Expected value and variance
c. given he looked in two stores and didn't find the product, what's the probability he won't find it at all?

About A: If he looked in $k=1,...,5$ stores, he didn't find the product in k-1 stores and found it in the last store so the probability is $\frac{1}{4}(\frac 3 4)^{k-1}$, so the distribution is 
but since the sum of probabilities is 1, it follows that $P(X=0)=0.501$. Why this result is intuitively correct.
About B:I found just by looking at the table that $E[X]=0.8271,Var(X)=0.9984$.
About C:if we define $B\text{-number of stores he  looked and didn't find it}$ we are lookign for $P(B=5\mid X=2)=\frac{P(B=5,X=2)}{P(X=2)}=\frac{P(B=5)}{P(X=2)}=\frac{(\frac 3 4) ^5}{\frac 3 {16}}=1.2656$ which is obviously wrong.
How can I explain/find $P(X=0)$ and why C is not correct?
 A: If we go into a store, let $S$ stand for success, the store had it, and let $F$ stand for failure. Note that the probability of success is $\frac{1}{4}$ and the probabiliy of failure is $\frac{3}{4}$
As you had it, $\Pr(X=1)=\Pr(S)=\frac{1}{4}$. 
$\Pr(X=2)=\Pr(FS)=\frac{3}{4}\cdot \frac{1}{4}=\frac{3}{16}$.
$\Pr(X=3)=\Pr(FFS)=\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{1}{4}=\frac{9}{64}$.
$\Pr(X=4)=\Pr(FFFS)=\frac{27}{256}$.
For $\Pr(X=5)$ things are a little different, because of the "give up" condition. There are two ways to find $\Pr(X=5)$. We could add up the probabilities previously obtained, and subtract the sum from $1$. Or else we can note that $X=5$ if we have four failures in a row. So $\Pr(X=5)=\frac{81}{256}$.
Mean and variance are somewhat tedious calculations. I will assume that, now that you have the distribution, you can carry them out. 
We turn to the conditional probability problem. First we do it mechanically.
Let $A$ be the event "looked without success in two stores" and $B$ the event "won't find." We want $\Pr)B|A)$, which by definition is
$$\frac{\Pr(A\cap B)}{\Pr(A)}.$$
The top is just the probability of $FFFFF$. This is $(3/4)^5$. The bpttom is the probability of $FF$. This is $(3/4)^2$. Divide. We get $(3/4)^3$. 
Remark: The answer to the conditional probability problem is clear without all the symbols. Given that we failed twice, the probability we will strike out completely is $(3/4)^3$, since we now are going only to three stores.
