Expected value of q given y is weighted average of mean q and and y It is assumed that:  
1) $y=q+u$  
Where $q$ is productivity and $y$ a testscore that measures true productivity. $u$ is a normally distributed error term, independent of $q$, with zero mean and constant variance; $q$ is also assumed to be normally distributed with a mean equal to $\alpha$ and with a constant variance. The outcome of this is:  
2) $E(q | y) = (1-\gamma)\alpha + \gamma y$
where $\gamma=Var(q)/(Var(q)+Var(u))$
How do you get equation 2? The equation can be expressed as a group effect and a individual effect.  
[This is a model of statistical discrimination, see:
Dennis J. Aigner and Glen G. Cain. Statistical theories of discrimination in labor markets.
Industrial and Labor Relations Review, 30(2):175{187, January 1977. URL http://
ideas.repec.org/a/ilr/articl/v30y1977i2p175-187.html.]
 A: All you need to know is that the regression of $q$ on $y$ is determined by standardizing both variables and their correlation coefficient will be the slope.
(In particular this result owes nothing to the assumptions that distributions are Normal; the independence of $q$ and $u$ is sufficient.  Thus it will be most revealing to obtain it without recourse to any properties of Normal distributions.)

Preliminary Calculations
To standardize a variable, you subtract its expectation and divide by its standard deviation.  We will therefore need to compute standard deviations, expectations, and a correlation coefficient.
Because $y=q+u$,
$$\mathbb{E}(y) = \mathbb{E}(q+u) = \mathbb{E}(q) + \mathbb{E}(u) = \alpha + 0 = \alpha,$$
taking care of computing the expectations.
Turn now to the standard deviations.  Recall that it's simpler to work with their squares: the variances.  For brevity, write $\sigma^2$ for the variance of $q$ and $\tau^2$ for the variance of $u$.  Then
$$\text{Var}(y) = \text{Var}(q+u) = \text{Var}(q) + \text{Var}(u) + 2\text{Cov}(u,q) = \sigma^2 + \tau^2 + 0 = \sigma^2 + \tau^2.$$
Finally, the correlation is computed from the covariance:
$$\text{Cov}(y, q) = \text{Cov}(q+u, q) = \text{Cov}(q,q) + \text{Cov}(u,q) = \sigma^2.$$
(Both these calculations used the simplification $\text{Cov}(u,q)=0$ arising from the independence of $u$ and $q$.)
Therefore the standardized variables are $$\eta = (y-\alpha)/\sqrt{\sigma^2+\tau^2}$$ and $$\theta=(q-\alpha)/\sigma.$$
Moreover, the correlation is $$\rho=\sigma^2/\left(\sigma\sqrt{\sigma^2+\tau^2}\right) = \sigma / \sqrt{\sigma^2+\tau^2}.$$

Solution
We have computed everything necessary to regress $q$ against $y$:
$$\mathbb{E}(\theta\ |\ \eta) = \rho\, \eta.$$
(This is a fact about geometry, really: see the "Conclusions" section at https://stats.stackexchange.com/a/71303 for the derivation, which--although it is illustrated there for Normal distributions--still does not require Normality to derive.)
Expanding, and once again exploiting linearity of expectation,
$$\frac{\mathbb{E}(q\ |\ y)-\alpha}{\sigma} = \mathbb{E}(\theta\ |\ \eta) = \rho\, \eta = \frac{\sigma}{\sqrt{\sigma^2+\tau^2}}\left(\frac{y-\alpha}{\sqrt{\sigma^2+\tau^2}}\right) = \frac{\sigma(y-\alpha)}{\sigma^2+\tau^2}.$$
It is the task of ordinary algebra to convert this back to an expression for $\mathbb{E}(q\ |\ y)$ in terms of $y$, because (insofar as $\mathbb{E}(q\ |\ y)$ is concerned) all variables now represent numbers:
$$\mathbb{E}(q\ |\ y) = \frac{\tau^2}{\sigma^2+\tau^2} \alpha + \frac{\sigma^2}{\sigma^2+\tau^2} y.$$
That is Equation (2).  Casting an eye back over the calculations should relieve any mystery about where these coefficients came from or what they mean.
A: The model implies that $y\sim\mathcal{N}(q,\sigma^2_u)$ and $q\sim\mathcal{N}(a,\sigma^2_q)$. By Bayes' rule:
$$p(q\mid y)\propto p(y\mid q,\sigma^2_u)p(q)$$
Ignoring constant factors (see here for a similar development):
$$\begin{align}p(q\mid y) & \propto \exp\left\{-\frac{(y-q)^2}{2\sigma^2_u}-\frac{(q-a)^2}{\sigma^2_q}\right\}\\
&=\exp\left\{-\frac{1}{2}\left(\frac{y^2-2yq+q^2}{\sigma^2_u}+\frac{q^2-2qa+a^2}{\sigma^2_q}\right)\right\}\end{align}$$
any term that does not include $q$ can be viewed as a proportionality constant:
$$\begin{align}\qquad\qquad\qquad &\propto\exp\left\{-\frac{1}{2}\frac{-2\sigma^2_q yq+\sigma^2_q q^2+\sigma^2_u q^2-2\sigma^2_u qa}{\sigma^2_u\sigma^2_q}\right\}\\
&=\exp\left\{-\frac{1}{2}\frac{(\sigma^2_q+\sigma^2_u)q^2-2(\sigma^2_u a+\sigma^2_q y)q}{\sigma^2_u\sigma^2_q}\right\}\\
&=\exp\left\{-\frac{1}{2}\frac{q^2-2q\frac{\sigma^2_u a+\sigma^2_q y}{\sigma^2_q+\sigma^2_u}}{\frac{\sigma^2_q\sigma^2_u}{\sigma^2_q+\sigma^2_u}}\right\}\propto \exp\left\{-\frac{1}{2}\frac{\left(q-\frac{\sigma^2_u a+\sigma^2_q y}{\sigma^2_q+\sigma^2_u}\right)^2}{\frac{\sigma^2_q\sigma^2_u}{\sigma^2_q+\sigma^2_u}}\right\}\end{align}$$
Therefore:
$$E(q\mid y)=\frac{\sigma^2_u a+\sigma^2_q y}{\sigma^2_q+\sigma^2_u}
=\left(1-\frac{\sigma^2_q}{\sigma^2_q+\sigma^2_u}\right)a+\frac{\sigma^2_q}{\sigma^2_q+\sigma^2_u}y$$
A: Another way, the shortest one ;-)
In general, if $X$ and $Y$ have a bivariate normal distribution, then (Anderson, Theorem 2.5.1):
$$E[X\mid Y]=E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}(Y-E[X])
=\left(1-\frac{\text{Cov}(X,Y)}{V[Y]}\right)E[X]+\frac{\text{Cov}(X,Y)}{V[Y]}Y$$
i.e. "expected value of X given Y is weighted average of mean X and Y" is a well-known result.
In your model $E[q]=a$,  $V[y]=\sigma^2_q+\sigma^2_u$ and $\text{Cov}(y,q)=\sigma^2_q$ (see whubner's answer), so:
$$E[q\mid y]=a+\frac{\sigma^2_q}{\sigma^2_q+\sigma^2_u}(y-a)=
\left(1-\frac{\sigma^2_q}{\sigma^2_q+\sigma^2_u}\right)a+\frac{\sigma^2_q}{\sigma^2_q+\sigma^2_u}y$$
A: I think the following argument shows why, unfortunately it's a messy. Much more elegant derivations are certainly out there somewhere as the linear Gaussian case is the best understood statistical model in existence. 
Anyway, we have that:


*

*U~N(0, sigma²)

*Q~N(alpha, beta^2)

*Y=Q+U.

*U and Q are independent.
Because a linear function of normal random variables is itself a normal random variable and due to independence of U and Q it follows that Y | Q ~ N(Q+0,sigma²).
We can now write down the probability density function of Q conditional on Y=y. By Bayes theorem that's:
(pdf of Q times * pdf of Y|Q)/(pdf of Y).
I won't write this out because it's very messy with all the Gaussian densities. 
Q|Y will be a normal random variable, which means that it's mode is its mean. Ignoring the denominator (the normalizing constant), we're left with:
1/(2pi*sigma^2*beta^2) * exp(-Something(Q))
We find the mode of the posterior distribution by choosing Q so as to maximise the density. That's going to be the conditional expected value too, because the mode is the mean for a Gaussian. To do that we can ignore everything except Something(Q) because the rest isn't a function of Q.
If you do the algebra, Something(Q) = 1/2 *( (y-q)^2/sigma^2) + (q-alpha)^2/beta^2)
If you differentiate wrt q, set to 0 and solve for q, you'll get:
q=beta^2/(beta^2+sigma^2)*y+sigma^2/(beta^2+sigma^2)*alpha... as required! 
