1) If we let the lengths of the legs be x and 2x, then by the Pythagorean theorem,
[tex]x+2x+\sqrt{x^{2}+(2x)^{2}}=28\\3x+x\sqrt{5}=28\\x(3+\sqrt{5})=28\\x=\frac{28}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}=21-7\sqrt{5}[/tex]
This means that the length of the hypotenuse is:
[tex]\sqrt{5}(21-7\sqrt{5})=\boxed{21\sqrt{5}-35}[/tex]
2) (Diagram attached) We know that
[tex]\cos 51^{\circ}=\frac{x}{58}\\x=58 \cos 51^{\circ} \approx \boxed{36.5}[/tex] (in cm)
3) From the Pythagorean identity,
[tex]\sin^{2}\left (\arccos \frac{5}{13} \right)+\cos^{2} \left (\arccos \frac{5}{13} \right)=1\\\left(\frac{5}{13} \right)^{2}+\sin^{2} \left (\arccos \frac{5}{13} \right)=1\\\frac{25}{169}+\sin^{2} \left (\arccos \frac{5}{13} \right)=1\\\sin^{2} \left (\arccos \frac{5}{13} \right)=\frac{144}{169}\\\sin \left (\arccos \frac{5}{13} \right)=\boxed{\frac{12}{13}}[/tex]
4) [tex]\frac{\pi}{6}[/tex]
