A bag contains 7 blue balls and 3 green balls. What is the probability of the following events? A bag   contains    7   blue    balls   and 3   green   balls.
I am not sure about my answer but please allow me to share.

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*If 5 balls are drawn in succession at random without replacement, what is the probability of drawing 1 green, 1 blue, 1 green, and 1 blue in order?

There are 10 balls in the bag initially, 7 of which are blue and 3 balls are green. All balls are equally likely to be drawn from the bag.
Since the order is important, we need to compute the probability of getting 1 green, 1 blue, 1 green, and 1 blue in order.
Hence,
$GBGB = (\frac{3}{10})(\frac{7}{9})(\frac{2}{8})(\frac{6}{7}) = \frac{1}{20}$
The probability of getting 1 green, 1 blue, 1 green, and 1 blue in order is $\frac{1}{20}$.


*If 3 balls are drawn at the same time, what is the probability that all the 3 balls drawn are blue?

$BBB = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = (\frac{7}{24})$


*If 5 balls are drawn at the same time, what is the probability that 3 balls are blue and 2 balls are green?

Work it out first where the order they are drawn is important.
$BBBGG = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8})(\frac{3}{7})(\frac{2}{6}) = (\frac{1}{24})$
Then work out possible different orders they could be drawn, there are 20 permutations $P(5,2)$. so the answer is
$20 × (\frac{1}{24}) = (\frac{5}{6})$
Any comments or suggestions will be much appreciated. Thank you in advance.

Problem 1.
Your answer to problem 1. is wrong, because you have computed the probability
of getting G-B-G-B in the first four drawings.  The problem specifically
asked for the probability of the satisfying sequence occuring anywhere
within the first five drawings.
The actual problem is much more complicated, and is best attacked by the
following approach:  the answer will be
$$\frac{N\text{(umerator)}}{D\text{(enoninator)}}$$
where
$$D = 10 \times 9 \times 8 \times 7 \times 6.$$
Because of the constraints of the problem, the easiest approach is to identify
each of the 4 possible satisfying sequences, and enumerate each possibility.
$N_1~: G-B-G-B-G ~:~N_1 = 3 \times 7 \times 2 \times 6 \times 1.$
$N_2~: G-B-G-B-B ~:~N_2 = 3 \times 7 \times 2 \times 6 \times 5.$
$N_3~: G-G-B-G-B ~:~N_3 = 3 \times 2 \times 7 \times 1 \times 6.$
$N_4~: B-G-B-G-B ~:~N_4 = 7 \times 3 \times 6 \times 2 \times 5.$
Final answer:
$$ \frac{N_1 + N_2 + N_3 + N_4}{D}.$$

Problem 2.
Your answer is correct.

Problem 3.
Your answer is almost right.  You overlooked overcounting.
There are only $\binom{5}{2} = 10$ ways of selecting which slots the 2 blue
balls will go in.  So you need to apply a scaling factor of (1/2) to your answer.
