Question: I am looking for an example of two random variables $X,Y$ such that 

(a) $X,Y$ are not independent.

(b) At least one of $X,Y$ is not normal.

(c) $E(X|y)$ (expected value of $X$ given $Y=y$) is linear in $y$, i.e. of the form $a+by$, and $E(Y|x)$ is linear in $x$.

(d) The correlation coefficient $\rho\neq \pm 1$.

Answer: 
A general method to get such $X, Y$ is as follows: 

Let $Y,Z$ be independent identically distributed random variables. Let $X=Y+Z$. Then as noted above $E(X|y)=y+E(Z)$ is linear in $y$. 

And intuitively, $E(Y|x)=E(Z|x)=x/2$. This can be proved rigorously (for continuous case) as follows:

Suppose the density function for $Y,Z$ is $f(u)$. Then the joint pdf for $X,Y$ is $f(y)f(x-y)$, and the density function for $X$ is the convolution of $f$ with itself. Then
$$ E(Y|x)=\frac{\int y f(y)f(x-y) dy}{\int f(y)f(x-y) dy} .$$ 
This is equal to $x/2$ as shown by Eric Naslund in [this post].

[Proving an integral identity: $\int\nolimits\_{-\infty}^\infty x f(x)f(t-x) dx =\frac{t}{2} \int\_{-\infty}^\infty f(x)f(t-x) dx $](https://math.stackexchange.com/questions/33997/an-integral-identity)
$Y, Z$ are r.v.'s. Let $X=aY+bZ$. Then the correlation should not be $\pm1$.