The given system of equations is
\((\cos^{-1}x)^2 \sin^{-1}y = 1\) and \(\cos^{-1}x (\sin^{-1}y)^2 = 1\).


To solve the system of equations, we can start by rearranging the first equation as follows:

\((\cos^{-1}x)^2 \sin^{-1}y = 1\)
\(\Rightarrow (\cos^{-1}x)^2 = \frac{1}{\sin^{-1}y}\)
\(\Rightarrow \cos^{-1}x = \sqrt{\frac{1}{\sin^{-1}y}}\)
\(\Rightarrow x = \cos\left(\sqrt{\frac{1}{\sin^{-1}y}}\right)\) .....(1)

Similarly, we can rearrange the second equation as follows:

\(\cos^{-1}x (\sin^{-1}y)^2 = 1\)
\(\Rightarrow \cos^{-1}x = \frac{1}{(\sin^{-1}y)^2}\)
\(\Rightarrow x = \cos\left(\frac{1}{(\sin^{-1}y)^2}\right)\) .....(2)

Now, let's compare equations (1) and (2) to find the possible values of x and y.

Case 1: If \(\sqrt{\frac{1}{\sin^{-1}y}} = \frac{1}{(\sin^{-1}y)^2}\), then x will have a unique value.

Squaring both sides of the equation, we get:
\(\frac{1}{\sin^{-1}y} = \frac{1}{(\sin^{-1}y)^4}\)
\((\sin^{-1}y)^4 - (\sin^{-1}y) = 0\)
\((\sin^{-1}y)((\sin^{-1}y)^3 - 1) = 0\)

Since \(\sin^{-1}y\) cannot be zero, we have:
\((\sin^{-1}y)^3 - 1 = 0\)
\((\sin^{-1}y)^3 = 1\)
\(\sin^{-1}y = 1\)

Therefore, \(y = \sin(1)\) and from equation (1), \(x = \cos\left(\sqrt{\frac{1}{\sin^{-1}y}}\right) = \cos(1)\).

Case 2: If \(\sqrt{\frac{1}{\sin^{-1}y}} \neq \frac{1}{(\sin^{-1}y)^2}\), then x and y will have no solution in this case.

Therefore, the system of equations has only one solution:
(x, y) = \((\cos(1), \sin(1))\).

Thus, the correct answer is '1'.