As follows:
I. Melody has shown that the result is pi/4. This is 0.79 to two dp.
II. Use the fact that \(e^{i\pi/2}=i\) so \(i^i=(e^{i\pi/2})^i\rightarrow e^{i\times i\pi/2}\rightarrow e^{-\pi/2}\)
Hence \(\chi=5+ie^{-\pi/2}\)
\(Real(\chi)=5.00\) to two dp and \(Imag(\chi)=0.21\)to two dp
III. Separate variables and use the fact that \(\frac{1}{\psi(1-\psi)}=\frac{1}{\psi}+\frac{1}{1-\psi}\)the integrals are then straightforward. Remember to include a constant, k, say, because the integrals are indefinite. You then have two unknowns, \(\beta\) and k. Use the two given conditions to find them. Then you can determine \(\psi(4)\). (You should find \(\psi(4) = 0.99\) to two dp).
(I would have supplied more detail, but the image uploading process isn't working at present!)
+33665 
