Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:

Given  $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.

The first couple of steps are the same: finding  $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}} 
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65}   \\
\sin x &=\frac {8} {\sqrt{65}} 
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2} 
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$  OR  $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant. 
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$ 
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $   OR   $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.

Simply: Solving for an angle using its sine value is ambiguous. (This is one reason that the Law of Cosines is preferred over the Law of Sines for determining angles from sides.) 

Here's an analogous situation:
$$\begin{align}x^2 + 2 x = 35 \quad(1)\\
x^2 - 2 x = 15 \quad(2)
\end{align}$$
Adding $(1)+(2)$, we see that $2 x^2 = 50$, so that $x = \pm 5$. On the other hand, $(1)-(2)$ gives $4x = 20$, so that $x = 5$. Two valid approaches seem to give different solution sets. But note: in the first case, one of the proposed values of $x$ is extraneous. You can check this by substitution: $-5$ is not a solution to $(1)$ or $(2)$.

In your problem, you can likewise determine that one of the values from the sine-based solution is extraneous.
We're given that $\cos 2x$ is negative, while $x$ itself is strictly positive and acute; therefore, $2x$ must be strictly larger than a right angle, which implies that $x > \pi/4$. Since $y$ is also assumed positive, it cannot be that $x+y = \pi/4$. Thus, $3\pi/4$ is the only valid solution.

The lesson here is that some approaches to a problem introduce extraneous solutions. You should always double-check that proposed solutions actually work in the original equations.
