# 6.1

From the minimality, it follows that $n-1=m^{2}$ and, by the condition, $n+2015<(m+1)^{2}$. Therefore, $2016<(m+1)^{2}-m^{2}=2 m+1$. From this, $m \geq 1008$. It remains to note that $n=1008^{2}+1$ fits.

№13

Since the coefficient of $x^{2}$ is positive for any value of the parameter $a$, the condition of the problem is equivalent to the specified function taking a negative value at the point $x=1$. Hence, $a \in(-2-\sqrt{11} ;-2+\sqrt{11})$.
Answer. $a \in(-2-\sqrt{11} ;-2+\sqrt{11})$.

#14

Subtracting the second equation from the first, we get: $\sin \alpha-\cos \alpha-\cos (\beta+\gamma)=\frac{1}{2}$, which means $\sin \alpha-\cos \alpha-\cos (\pi-\alpha)=\sin \alpha=\frac{1}{2}$. There are two possible cases:

1) $\alpha=30^{\circ}$. Then $\sin \beta \sin \gamma=\frac{\sqrt{3}}{4}, \cos \beta \cos \gamma=-\frac{\sqrt{3}}{4}$. From this, $\cos (\beta-\gamma)=0$, which means $\beta-\gamma= \pm 90^{\circ}$. But $\beta+\gamma=150^{\circ}$. Therefore, $\{\beta, \gamma\}=\left\{30^{\circ}, 120^{\circ}\right\}$
2) $\alpha=150^{\circ}$. Then $\cos \beta \cos \gamma=\frac{3 \sqrt{3}}{4}>1$, which is impossible.

Answer. $30^{\circ}, 30^{\circ}, 120^{\circ}$.

## #15

From the fact that $\angle E C F=90^{\circ}$, it follows that $E F$ is the diameter of the circle and $C G=\frac{1}{2} C D$. Let $\alpha=\angle A$, then $\angle A C D=90^{\circ}-\alpha \Rightarrow C E=C D \cdot \cos \left(90^{\circ}-\alpha\right)=C D \sin \alpha$. Similarly, $\angle B=90^{\circ}-\alpha \Rightarrow \angle B C D=\alpha \Rightarrow C F=C D \cos \alpha$. From the given ratio in the problem, we get: $\frac{1}{4} C D^{2}=C D^{2} \cos \alpha \cdot \sin \alpha$, i.e., $\frac{1}{2}=\sin 2 \alpha, 2 \alpha=30^{\circ}$.

Answer. $15^{\circ}, 75^{\circ}, 90^{\circ}$.