A person has two children. What is the probability that both children are boys ,given that:
a) The first children is a boy
b)at least one of them is a boy.
What I did is that I've translated the a part to:
$P[(A\cap B) / A]$. and considered that $P(A)=0.5$
Where, $A$ is the event that the first child is a boy. $B$ is the event that the second is a boy.
For the b part , I translated it into:
$P[(A\cap B)/(A\cup B)]$. And after some simplifications I found that it equals $\frac{P(A\cap B)}{P(A\cup B)}$. And I computed the probability of the union by the addition rule and got the answer.
Is that methodology right? If not, what is the correct way to solve this problem?
Yes, you have it.
For (a) you want to find, as you note: $$\mathsf P(A\cap B\mid A) = \frac{\mathsf P(A\cap B)}{\mathsf P(A)}\qquad\color{green}\checkmark$$
The event that each child is a boy is independent of that of the other, so the multiplication rule takes the form of: $$\mathsf P(A \cap B) = \mathsf P(A)\;\mathsf P(B)$$
Likewise for (b) you do want: $$\mathsf P(A\cap B\mid A\cup B) = \frac{\mathsf P(A\cap B)}{\mathsf P(A\cup B)}\qquad\color{green}\checkmark$$
Now, the events that either child is a boy are not mutually exclusive, so the addition rule takes the form (often known as the Principle of Inclusion and Exclusion) of: $$\mathsf P(A\cup B) = \mathsf P(A)+ \mathsf P(B) - \mathsf P(A\cap B)$$
Just use $\mathsf P(A)=\mathsf P(B) \approx \frac 1 2$ and you're done.
Remark Another method is to simply list the possible outcomes
$$\{\mathop{\mathsf{\underbrace{\overbrace{BB, BG}, GB}, GG}}^\text{First child a boy}_\text{At least one child is a boy}\}$$
Since all outcomes are equally likely, we directly obtain the results: $$\begin{align}\mathsf P(A\cap B\mid A) & =\frac \Box \Box \\[1ex] \mathsf P(A\cap B\mid A\cup B) & = \frac \Box \Box\end{align}$$