Why does aluminium-on-glass mirror work without distortion? I have read an article about glass (zerodur) with low thermal expansion coefficient. It is mentioned that large casts of such glass are covered with reflective layer of Aluminium and used as mirrors in space observatories.
Low CTE is so important in this glass, because changes in size of it would distort the picture taken with the telescope. But what about said layer of aluminium? It is metal and it's CTE is much larger than the one of the glass.
So how does it happen that the thin layer of aluminium doesn't distort the picture? Doesn't it expand/shrink? 
 A: Let us take the example of the Hubble primary mirror. It has a diameter of 2.4 m and a mass of 828 kg. It is actually made in a sandwich structure - glass-honeycomb-glass - making it about 30 cm thick (for stiffness) but light.
The mirror is coated with an aluminum coating of thickness t = 65 nm, with a 25 nm MgF2 protective coating on top.
Coefficient of thermal expansion of aluminum is $2.2 \cdot 10^{-5} \mathrm{m/m\cdot K}$ and the Young's modulus is 69 GPa. If you constrain the coating to be of constant size, then you increase the strain by $2.2 \cdot 10^{-5}$ per °C, and for a square sheet of aluminum with sides $L$ and thickness $t$, the force this creates would be
$$F = \sigma E A = 2.2\cdot 10^{-5} \cdot 69 \cdot 10^9 \cdot 2.4 \cdot 10^{-9} \approx 0.4 N $$
A force of 0.4 N across a 30 cm thick mirror gives a bending moment of about 0.06 Nm; the radius of curvature can be approximated by
$$R = \frac{EI}{M}$$
Again, we will approximate the mirror as a square with 1 D deflection, in which case the second moment of area is given by approximately (t = thickness of glass = 3.5 cm, s = spacing = 25 cm)
$$I = \frac{Lt}{2 s^2} \approx 0.001 m^4$$
Thus we find for the (additional) curvature of the mirror:
$$R = \frac{70\cdot 10^9 \cdot 0.001}{0.4}=3\cdot 10^9 m$$
This results in a deflection across the 2.4 m diameter of 
$$\Delta h = \frac{r^2}{2R} = \frac{1.44}{2\cdot 3\cdot 10^9} \approx 0.3 \mathrm{nm}$$
For comparison it is worth noting that the specification of the mirror calls for the shape to be accurate to within 10 nm, and the error that led to the initial blurry images was on the order of 2 µm. In other words, while the coating might have an effect, it is quite a bit smaller than the specification of the mirror.
And there is another thing - the mirror has a heater on it which keeps the temperature constant within half a degree (to maintain shape). Originally this temperature was 21 degrees, but that makes it work less well in the near IR. Later the temperature was dialed down to 15°C (source - as above).
So yes - thermal expansion of coatings can affect the shape of the mirror; but no, it is not significant.
