Since $\displaystyle c=90^\circ$, we know we are working with a right triangle.

That means that $\displaystyle \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$.

In this problem, that would be:

$\displaystyle \sin(a)=\frac{\text{Z}}{\text{X}}$

Plug in our given values:

$\displaystyle \sin(a)=\frac{8}{12}$

$\displaystyle \sin(a)=\frac{2}{3}$

$\displaystyle a=\sin^{-1}(\frac{2}{3})$

$\displaystyle a=41.8^\circ$

$\displaystyle sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$

$\displaystyle \Rightarrow sin^2x+(cos^2x+cos^2x)+3(sin\ x)(cos\ x)=3$

Use the identity $\displaystyle sin^2x+cos^2x=1$ .  

$\displaystyle \Rightarrow 1+cos^2x+3(sin\ x)(cos\ x)=3\Rightarrow cos^2x+3(sin\ x)(cos\ x)=3-1=2$

Now we should divide both sides by $\displaystyle sin^2x$:

$\displaystyle \Rightarrow \frac{cos^2x}{sin^2x}+\frac{3(sin\ x)(cos\ x)}{sin^2x}=\frac{2}{sin^2x}$

We can use the identity   $\displaystyle \frac{1}{sin^2x}=1+cot^2x$.

$\displaystyle \Rightarrow cot^2x+3cot\ x=2(1+cot^2x)$

$\displaystyle \Rightarrow cot^2x+3cot\ x=2+2cot^2x$

$\displaystyle \Rightarrow 2cot^2x-cot^2x-3cot\ x+2=0$

$\displaystyle \Rightarrow cot^2x-3cot\ x+2=0$

$\displaystyle \Rightarrow (cot\ x-1)(cot\ x-2)=0$
 

$\displaystyle cot\ x-1=0\Rightarrow cot\ x=1$

or

$\displaystyle cot\ x-2=0\Rightarrow cot\ x=2$

$\displaystyle sin\ x+\frac{1}{sin\ x}=2\Rightarrow \frac{sin^2x+1}{sin\ x}=2$

$\displaystyle \Rightarrow sin^2x+1=2sin\ x$

$\displaystyle \Rightarrow sin^2x-2sin\ x+1$

$\displaystyle \Rightarrow (sinx-1)^2=0$

$\displaystyle \Rightarrow sin\ x-1=0$

$\displaystyle \Rightarrow sin\ x=1$

Therefore, $\displaystyle sin^5x+sin^7x=1^5+1^7=1+1=2$.

$\displaystyle x=\frac{\pi}{6}\Rightarrow \frac{{1+cos\ x}}{}{sin^3x}=\frac{{1+cos\ (\pi/6)}}{sin^3(\pi/6)}=\frac{1+\frac{\sqrt{3}}{2}}{(\frac{1}{2})^3}=4(2+\sqrt{3})=8+4\sqrt{3}$

We know that $\displaystyle sec\ x= \frac{1}{cos\ x}$.
 

$\displaystyle sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}=sin^2 (\frac{\pi}{15})+\frac{1}{\frac{1}{cos^2(\frac{\pi}{15})}}=sin^2 (\frac{\pi}{15})+cos^2 (\frac{\pi}{15})=1$

We need to use the identity $\displaystyle sec^2x=1+tan^2x$:

$\displaystyle sec^2x=1+tan^2x\Rightarrow (\frac{\sqrt{5}}2)^2=1+tan^2x$

$\displaystyle \Rightarrow tan^2x=\frac{5}{4}-1\Rightarrow tan^2x=\frac{1}{4}$

$\displaystyle \Rightarrow tan\ x=\pm \frac{1}{2}$

We need to use the identity $\displaystyle csc(90^{\circ}-x)=sec\ x$:

$\displaystyle sec\ 26^{\circ} - csc\ 64^{\circ}=sec\ 26^{\circ} - csc (90-26)^{\circ}=sec\ 26^{\circ} -sec\ 26^{\circ}=0$

We need to use the identity $\displaystyle cos(90^{\circ}-x)=sin\ x$:

$\displaystyle \frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}=\frac{cos (90-25)^{\circ}}{sin\ 25^{\circ}}=\frac{sin\ 25^{\circ}}{sin\ 25^{\circ}}=1$

We need to use the identity $\displaystyle cos2x=2cos^2x-1$.

$\displaystyle cos2x=2cos^2x-1=2A^2-1$