Let f(x) = 3 + x^2 + tan(pi/(2x)) , -1 < x < 1 a) Find f^-1(3) b) Find f(f^-1 (5))

Hello!
You wrote f(x) = 3+x^2+tan(pi/(2x)).
By the definition of an inverse function of f(x),  f^(-1)(3) is that number x for which f(x) = 3. Usually we require that such a number must be unique, otherwise f^(-1) would be a many-valued function.
a. In other words, we need to solve the equation f(x) = 3.
In our problem, f(x) takes any value infinitely many times, even at the given interval (-1, 1), even at any neighborhood of x = 0.
The cause of this is that tan(pi/(2x)) tends to +-oo at points where pi/(2x) = pi/2 + k pi for some integer k. The 3+x^2 part remains finite and bounded at any finite interval and cannot prevent this behavior of f(x). These points are x_k = 1/(1+2k) and they tend to zero as k tends to +-oo.
Regardless of the number of solutions, the equation f(x)=3, which is equivalent to x^2+tan(pi/(2x)) = 0, cannot be solved exactly.
I might suppose that you misprint the formula, probably f(x) = 3+x^2+tan(pi/2 x). In that case, the only solution for f(x)=3 at the interval (-1,1) is x=0. This is because f is strictly monotone on (-1,1). It is not obvious but true. Ask me if you need a proof.
b. If f^(-1)(5) exists, then by definition f(f^(-1)(5)) = 5.