$a)\sin \left(\frac{7}{2}\pi +x\right)=-\cos \left(x\right)=-\left(-\frac{5}{13}\right)=\frac{5}{13}$  

 

$b)\cos \left(5\pi +x\right)=-\cos \left(x\right)=-\left(-\frac{5}{13}\right)=\frac{5}{13}$  

 

$c)\sin \left(x-\frac{7}{2}\pi \right)=-\sin \left(\frac{7}{2}\pi -x\right)=-\left(-\cos x\right)=\cos x=-\frac{5}{13}$  

 

$d)\cos \left(x-\frac{5}{2}\pi \right)=\cos \left(-\left(\frac{5}{2}\pi -x\right)\right)=\cos \left(\frac{5}{2}\pi -x\right)=\sin x=\frac{12}{13}$  

${\sin }^{2}x=1-{\cos }^{2}x$    

${\sin }^{2}x=1-\frac{25}{169}$  

${\sin }^{2}x=\frac{144}{169}$  

$\left|\sin x\right|=\frac{12}{13}$  

$\text{Dla}x\in \left(\frac{\pi }{2},\pi \right)\sin x>0$  

$\sin x=\frac{12}{13}$  

 

$e)\mathrm{tg}\left(\frac{11}{2}\pi +x\right)=-\mathrm{ctg}\left(x\right)=-\frac{\cos x}{\sin x}=-\frac{-\frac{5}{13}}{\frac{12}{13}}=\frac{5}{13}\cdot \frac{13}{12}=\frac{5}{12}$  

 

$f)\mathrm{ctg}\left(-\left(\frac{13}{2}\pi -x\right)\right)=-\mathrm{ctg}\left(\frac{13}{2}\pi -x\right)=-\mathrm{tg}\left(x\right)=-\frac{1}{\mathrm{ctg}x}=\frac{1}{-\mathrm{ctg}x}=\frac{1}{\frac{5}{12}}=\frac{12}{5}$