Independent Probability Problem 
If we repeat a same experiment in a laboratory independently 6 times.
  Evey time the probability that the experiment succeeds is 0.7. 
a)What is the probability that it succeeds exactly 5 times?
b) What is the probability that it succeeds exactly twice?

 A: Anytime that you have situation involving a sequence of $n$ independent trials that result in either success or failure, and you want to know the probability of exactly $k$ successes, you have what is called a binomial distribution. The probability of exactly $k$ successes out of $n$ trials with probability $p$ of success is given by
$$
\begin{pmatrix} n\\k \end{pmatrix}p^k\left(1-p\right)^{n-k}.
$$
Where $(~^n_k)$, read as "n choose k," is the number of ways that you can choose $k$ elements from a set of $n$ elements and equals 
$$\frac{n!}{k!(n-k)!}.$$ 
In your case, $n=6$, $p=0.7$, and for part (a) $k=5$ and for part (b) $k=2$.
The logic for this construction is as follows. I will describe the process for part (a), and the logic will be the same for part (b). We can calculate the probability $P(SSSSSF)=(0.7)^5 0.3$ using the multiplication rule for independent events, which is the probability of getting first five successes and then one failure, but this is not the only way get five successes. We could also get the series $(SSSFSS)$, so we need to add up the probability for all of these different possibilities that have the same probability.
The number of such events called $(~^6_5)$ which is the number of ways to choose five sucesses out of six trials. If you need help with knowing why this is true, you can ask in a comment and I will add it to my answer, or better yet search SE for an answer. We can therefore multiply $(0.7)^50.3$ and $(~^6_5)$ to get the answer.
A: We want the probability of exactly $5$ successes, and therefore $1$ failure, in $6$ trials.  
The $5$ successes and $1$ failure could happen in several orders. They are $SSSSSF$, $SSSSFS$, $SSSFSS$, $SSFSSS$, $SFSSSS$, and $FSSSSS$,
Calculate the probability of each and add them up. For example, the probability of the pattern $SSSSSF$ is $(0.7)(0.7)(0.7)(0.7)(0.7)(0.3)$. 
For (b), you will probably want to count the number of patterns more efficiently than by listing.
A: André has answered your question, but for the record: with independent events, an and translates to multiplication, an or translates to addition.
ex.: in a coin-toss scenario, the probability of getting 2 heads in 2 tosses (heads and heads) is $1/2\times 1/2=1/4$. The probability of getting either heads or tails (in one toss) is $1/2+1/2=1$, which makes perfect sense. 
By applying this to your exercise, you should be able to figure out the answers. (don't forget to include the failure probabilities as well as the success probabilities)
