1. Let $A=\sin^{-1}\left(\frac35\right),\quad B=\sin^{-1}\left(\frac5{13}\right),\quad C=\sin^{-1}\left(\frac{33}{65}\right).$ We need to find $\cos(A+B+C).$

2. Since $A,B,C\in\left[-\frac\pi2,\frac\pi2\right]$, their cosines are positive: $\cos A=\sqrt{1-\left(\frac35\right)^2}=\frac45,$ $\cos B=\sqrt{1-\left(\frac5{13}\right)^2}=\frac{12}{13},$ $\cos C=\sqrt{1-\left(\frac{33}{65}\right)^2}=\frac{56}{65}.$

3. First compute $\sin(A+B)$ and $\cos(A+B)$: $\sin(A+B)=\sin A\cos B+\cos A\sin B$ $=\frac35\cdot\frac{12}{13}+\frac45\cdot\frac5{13}$ $=\frac{36}{65}+\frac{20}{65}=\frac{56}{65}.$

   $\cos(A+B)=\cos A\cos B-\sin A\sin B$ $=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac5{13}$ $=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}.$

4. Now compute $\cos(A+B+C)$: $\cos(A+B+C)=\cos(A+B)\cos C-\sin(A+B)\sin C.$

   Substituting values, $\cos(A+B+C)=\frac{33}{65}\cdot\frac{56}{65}-\frac{56}{65}\cdot\frac{33}{65}=0.$

5. Therefore, $\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right)=0.$

6. Option check:

   • A: $\frac{33}{65}$ ❌
   • B: $1$ ❌
   • C: $\frac{32}{65}$ ❌
   • D: $0$ ✅

Hence, the correct answer is D.