Let $\alpha = \cot^{-1}(5), \qquad \beta = \cos^{-1}\left(\frac45\right).$ We need to find $\csc(2\alpha+\beta).$

First find trigonometric values for $\alpha$. Since $\cot \alpha = 5,$ we have $\tan \alpha = \frac{1}{5}.$ Take a right triangle with opposite $=1$, adjacent $=5$. Then hypotenuse is $\sqrt{1^2+5^2}=\sqrt{26}.$ So, $\sin \alpha = \frac{1}{\sqrt{26}}, \qquad \cos \alpha = \frac{5}{\sqrt{26}}.$

Now compute $\sin 2\alpha$ and $\cos 2\alpha$: $\sin 2\alpha = 2\sin\alpha\cos\alpha = 2\cdot \frac{1}{\sqrt{26}}\cdot \frac{5}{\sqrt{26}}=\frac{10}{26}=\frac{5}{13},$ $\cos 2\alpha = \cos^2\alpha-\sin^2\alpha = \frac{25}{26}-\frac{1}{26}=\frac{24}{26}=\frac{12}{13}.$

Now find trigonometric values for $\beta$. Given $\cos \beta = \frac45,$ and since $\beta=\cos^{-1}(4/5)$ lies in $[0,\pi]$, here it is acute, so $\sin \beta = \sqrt{1-\left(\frac45\right)^2} = \sqrt{1-\frac{16}{25}}=\frac35.$

Use the angle addition formula: $\sin(2\alpha+\beta)=\sin 2\alpha \cos\beta + \cos 2\alpha \sin\beta.$ Substitute values: $\sin(2\alpha+\beta)=\frac{5}{13}\cdot \frac45 + \frac{12}{13}\cdot \frac35.$ $=\frac{20}{65}+\frac{36}{65}=\frac{56}{65}.$

Therefore, $\csc(2\alpha+\beta)=\frac{1}{\sin(2\alpha+\beta)}=\frac{65}{56}.$

Comparing with the options, this is $\boxed{\frac{65}{56}}$ which is option B.