Your solutions to problem 1 and 2.1 are correct. For 2.2 you have calculated "dS" using the length of the normal vector at the single point \(\displaystyle (4, -2, 0)\) where u= 2 and \(\displaystyle v= \pi\). You can't do that. It has to be at any u, v. So \(\displaystyle dS= \sqrt{ u^2+ 4u^2cos^2(v)+ 4u^2sin^2(v)}dudv=\sqrt{5u^2}dudv= u\sqrt{5}dudv\). Yes, \(\displaystyle x(y^2+z^2)= 2u(u^2cos^2(v)+ u^2sin^2(v))= 2u(u^2)= 2u^3\) (it seems more work that necessary to do it as \(\displaystyle xy^2+ xz^2\)). So the integral is \(\displaystyle 2\sqrt{5}\int_0^{2\pi}\int_0^3 u^4 dudv\).
For problem 3 you have one notational errors, perhaps a "typo". You have "\(\displaystyle dS= \sqrt{14}\)" when it should be, of course, \(\displaystyle dS= \sqrt{14}dxdy\). But you do have the correct integral.
Problem 4 looks good.
