Why does $P(X+Y=0)=0$ imply that $X+Y$ is continuous 

In the answer to b) , i'm struggling to understand why is the conditional probability equal to 1 and the other is not. How do I calculate the conditional probability, since $P(X)=1/2$ is equal to $0$ since it's a continuous distribution?
In the answer to c) Why is it that $P(X+Y=0)=0$ implies that they are continuous and if $P(X+Y=0)=1$ implies that they are constant?  Following this, i'm again not sure how they calculate the probabilities $P(X+Y=0)=P( |X |>c)>0$ and that $P( |X |>c)$ is not equal to $1$ . 
 A: It does not. $X$ and $Y$ being continuous implies $P(X + Y = 0) = 0$. 
A: $\newcommand{P}{\mathbb P}\newcommand{E}{\mathbb E}$In general, 
suppose $P$ and $Q$ are two statements. The following sentences then all mean exactly the same thing:


*

*$P$ implies $Q$.

*If $P$ then $Q$.

*$Q$ if $P$.


All of these are very different in meaning from "$Q$ implies $P$."
For example:


*

*$T > 0$ if $T = 1$.


The statement above is a true statement for any real number $T$, but the following is not true for every real number $T$:


*

*$T>0$ implies $T=1$.


There is no place on the answer page where it says that $\P(X+Y=0)=0$ implies that $X+Y$ is continuous.
What it does say is this:

$\P(X+Y=0)=0$ if $X+Y$ is continuous

and moreover it only claims that this fact is true when $X+Y$ has a normal distribution.

Now let's consider what the answer does say.
For part (b), the answer asks you to find the probability
$\P\left(Y < c \mid X = \frac c2\right).$
Now we're already in trouble if you have not been given a suitable definition of what this conditional probability means.
Since $X$ is a continuous random variable, however, there are definitions that make sense; see the answers to
Probability, conditional on a zero probability event.
If you were given such a definition, it would presumably conform with the intuition that if $A \implies B$ and $\P(B\mid A)$ is defined, then $\P(B\mid A) = 1.$
And in this case $X=\frac c2 \implies Y=\frac c2$,
so $X=\frac c2 \implies Y\lt c.$
There is an alternative solution that does not require such sophistication:
$$\P\left(Y < c \;\middle|\; -\frac c2 \leq X \leq \frac c2\right) = 1,$$
because when $-\frac c2 \leq X \leq \frac c2$ then $Y=X$ and therefore
$-\frac c2 \leq Y \leq \frac c2$, so $Y < c$.
For part (c), the answer considers a general case where
$X+Y \sim \mathrm N(\mu,\sigma^2)$ where $\sigma^2\geq 0.$
The answer then breaks this down into two subcases:
Case $\sigma^2 > 0.$ Then $X+Y$ is a kind of normal distribution that we're accustomed to, with a continuous distribution with a non-zero density on every real number and a zero probability of being equal to any particular real number.
That is, $\P(X+Y=0) = 0.$
Case $\sigma^2 = 0.$ Then $X+Y = \mu,$ a constant.
Now you might already have determined from part (a) that $\E(Y) = 0.$
Hence $\mu = E(X+Y) = E(X) + E(Y) = 0,$ and therefore $X + Y = 0$ always,
so $\P(X + Y = 0) = 1.$
But the short version of the answer is that 
$\P(X + Y = 0) = \P(\lvert X\rvert > c)$ and therefore
$$ 0 < \P(X + Y = 0) < 1, $$
and there are no values of $\mu$ and $\sigma^2$ such that a random variable $Z$ distributed according to $Z \sim \mathrm N(\mu,\sigma^2)$ can have
$0 < \P(Z = 0) < 1.$

Why does $\P(X + Y = 0) = \P(\lvert X\rvert > c)$?
Remember the definition of $Y$ and consider two cases:
Case $\lvert X\rvert > c$. Then $Y = -X$ and $X + Y = 0.$
Case $\lvert X\rvert \leq c$. Then $Y = X$ and $X + Y = 2X,$
which is zero iff $X = 0$.
So the event that $X + Y = 0$ is exactly the same as the event:
$$ X = 0 \ \text{or}\ \lvert X\rvert > c. $$
Therefore $\P(X + Y = 0) = \P(X = 0 \ \text{or}\ \lvert X\rvert > c).$
Noting that $X=0$ and $\lvert X\rvert > c$ are disjoint events and that
$\P(X=0)=0,$
$$ \P(X = 0 \ \text{or}\ \lvert X\rvert > c)
= \P(X = 0) + \P(\lvert X\rvert > c)
= \P(\lvert X\rvert > c).$$
