1. Let

$$\alpha=\sin^{-1}\left(\frac45\right),\quad \beta=\sin^{-1}\left(\frac5{13}\right),\quad \gamma=\sin^{-1}\left(\frac{16}{65}\right).$$

We need to find

$$2\pi-(\alpha+\beta+\gamma).$$

2. Since all three values lie in $[0,1]$, their principal inverse-sine values are acute angles.

For $\alpha$:

$$\sin\alpha=\frac45 \implies \cos\alpha=\frac35.$$

Hence

$$\tan\alpha=\frac{4/5}{3/5}=\frac43.$$

For $\beta$:

$$\sin\beta=\frac5{13} \implies \cos\beta=\frac{12}{13}.$$

Hence

$$\tan\beta=\frac{5/13}{12/13}=\frac5{12}.$$

For $\gamma$:

$$\sin\gamma=\frac{16}{65} \implies \cos\gamma=\frac{63}{65}.$$

Hence

$$\tan\gamma=\frac{16/65}{63/65}=\frac{16}{63}.$$

3. First add $\alpha$ and $\beta$ using tangent addition:

$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} =\frac{\frac43+\frac5{12}}{1-\frac43\cdot\frac5{12}}.$$

Now simplify:

$$\frac43+\frac5{12}=\frac{16+5}{12}=\frac{21}{12}=\frac74,$$

and

$$1-\frac43\cdot\frac5{12}=1-\frac{20}{36}=1-\frac59=\frac49.$$

Therefore,

$$\tan(\alpha+\beta)=\frac{7/4}{4/9}=\frac74\cdot\frac94=\frac{63}{16}.$$

Since both $\alpha$ and $\beta$ are acute, $\alpha+\beta$ is acute as well, so

$$\tan(\alpha+\beta)=\frac{63}{16}.$$

4. Now compare with $\gamma$:

$$\tan\gamma=\frac{16}{63}.$$

Thus,

$$\tan(\alpha+\beta)\tan\gamma=1.$$

Also, both angles are acute, so

$$\alpha+\beta+\gamma=\frac\pi2.$$

(Indeed, if two acute angles have reciprocal tangents, they are complementary.)

5. Therefore,

$$2\pi-(\alpha+\beta+\gamma)=2\pi-\frac\pi2=\frac{3\pi}{2}.$$

6. Comparing with the options, the correct choice is:

$$\boxed{\frac{3\pi}{2}}$$

which is Option C.