Your first two pictures aren’t really helpful, so I’ve made better versions:

In the first picture $V$ is a neighborhood of the red point that does not contain any point not in $A$, so the red point is not a boundary point of $A$. In the second picture $V$ is a neighborhood of the red point that does not contain any point of $A$, so again the red point cannot be a boundary point of $A$. Only in your third picture is it true that every neighborhood of the red point must contain points of $A$ and points not in $A$, so it’s the only picture in which the red point is a boundary point of $A$.
The point $b+1$ is not a boundary point of $(a,b)$ because it has a neighborhood that does not contain any point of $(a,b)$. In fact it has many such neighborhoods, but one easy one is $\left(b+\frac12,b+2\right)$: $b+1\in\left(b+\frac12,b+2\right)$, but $\left(b+\frac12,b+2\right)\cap(a,b)=\varnothing$.
If $b=a+1$, then of course $a+1$ is a boundary point of $(a,b)$: every neighborhood of $b$ contains points less than $b$ that are in $(a,b)$ and points bigger than $b$ that are not in $(a,b)$. If $a+1, then $a+1\in(a,b)$, so $(a,b)$ itself is a neighborhood of $a+1$ that contains no points of $\Bbb R\setminus(a,b)$; this shows that $a+1$ is not a boundary point of $(a,b)$ in this case. If $a+1>b$, then $(b,a+2)$ is a neighborhood of $a+1$ that contains no points of $(a,b)$, and again $a+1$ is not a boundary point of $(a,b)$. (You’ll probably find it helpful to make drawings of these different cases.)