You can see that E[X - E(X|Y)]^2=1 by plugging in the 1 + 2(Y + 1)

E[X - E(X|Y)]^2=E[X - (2Y + 3)]^2 =E[X - 2Y - 3)]^2=V[X - 2Y - 3)]+ [E[X - 2Y - 3)]]^2

=V[X]+4V[Y] -4cov(X,Y)+ [E[X - 2Y - 3)]]^2

=5+4-8 +(0)^2=1

I was going to prove the first part from scratch but I knew it had to be somewhere online.

Look at.... Conditional distributions at ....http://en.wikipedia.org/wiki/Multiva...l_distribution

Plugging into that, you can see that 1 + 2(Y + 1) is correct

the 1 is the mean of X and the 2 is the ratio of the covariance over the variance of Y...