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Recall that

[tex]\cosh^2x-\sinh^2x=1\implies \cosh x=\pm\sqrt{1+\sinh^2x}[/tex]

You're given that [tex]\cosh x=\dfrac{13}5>0[/tex], so omit the negative root. Then

[tex]\dfrac{13}5=\sqrt{1+\sinh^2x}\implies\sinh^2x=\dfrac{144}{25}\implies\sinh x=\pm\dfrac{12}5[/tex]

Because [tex]x>0[/tex], you have [tex]\sinh x>0[/tex] too, so omit the negative root again. Then [tex]\sinh x=\dfrac{12}5[/tex].

Now,

[tex]\mbox{sech }x=\dfrac1{\cosh x}=\dfrac5{13}[/tex]
[tex]\mbox{csch }x=\dfrac1{\sinh x}=\dfrac5{12}[/tex]
[tex]\tanh x=\dfrac{\sinh x}{\cosh x}=\dfrac{\frac{12}5}{\frac{13}5}=\dfrac{12}{13}[/tex]
[tex]\coth x=\dfrac1{\tanh x}=\dfrac{13}{12}[/tex]

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