I'm very aware that you can use the CDF of the Gaussian distribution which would result in this:
Pr(-a<n<a)=Φ(a)-Φ(-a)=1-Q(a)-[1-Q(-a)]=1-Q(a)-[1-1+Q(a)]=1-2*Q(a)
I used the properties: 1.Φ(x)=1-Q(x) 2.Q(-x)=1-Q(x)
But I also tried to prove it with event algebra and failed miserably. I think my mistake is here: Pr(-a<n<a)=Pr(n>-a)*Pr(n<a)
so a little help would be appreciated!
**Edit: I've become aware Pr(-a<n<a)=1-Pr(n>2a) so the actual question is how does this work
For the first part of your question, we have $$ \begin{align*} \mathbb P(-a < N < a) & = \mathbb P( N < a \mbox{ and } N > -a) \\ & = 1- \mathbb P( N \geq a \mbox{ and } N \leq -a) \\ & = 1- \bigg(\mathbb P(N \geq a) + \mathbb P(N \leq -a)\bigg) \ \ \mbox{(events are disjoint)} \ (1) \\ & = \bigg(1 - \mathbb P(N \geq a)\bigg) - \mathbb P(N \leq -a) \\ & = \mathbb P(N \leq a) - \mathbb P(N \leq -a) \\ &= \phi(a) - \phi(-a) \end{align*} $$ where the main trick we use $\mathbb P(A) = 1-\mathbb P(A^C)$, and we also use that since $N$ is a continuous random variable, $\mathbb P(N \geq a) = \mathbb P(N > a)$ at the end/start depending on your definition.
To address the second question, I'm not sure whether what you wrote is accurate. But we can say this: looking at the line $(1)$, we multiply both sides by $(-1)$ then use the symmetry of the normal distribution about $0$ to say $$\mathbb P(N \geq a) = \mathbb P( -N \leq -a) = \mathbb P(N \leq -a)$$ so that $(1)$ reads exactly $$ \mathbb P(-a < N < a) = 1- \bigg(\mathbb P( N \leq -a) + \mathbb P(N \leq -a)\bigg) = 1 - 2 \cdot \mathbb P(N \leq -a) $$