$$\cos \left(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{33}{65}\right)$$ is equal to:

cos(sin⁻¹(3/5)+sin⁻¹(5/13)+sin⁻¹(33/65)).

sin⁻¹(3/5)+sin⁻¹(5/13): let α=sin⁻¹(3/5), β=sin⁻¹(5/13).

sin(α+β) = 3/5·12/13+4/5·5/13 = 36/65+20/65 = 56/65.

So α+β = sin⁻¹(56/65) (since cosα·cosβ>0).

Now sin⁻¹(56/65)+sin⁻¹(33/65): sin of sum = 56/65·√(1-1089/4225)+33/65·√(1-3136/4225)

= 56/65·√(3136/4225)+33/65·√(1089/4225) = 56·56/(65·65)+33·33/(65·65) = (3136+1089)/4225 = 4225/4225 = 1.

So the total = π/2. cos(π/2) = 0.

The correct answer is Option 2: 0.