Thevenin generator of circuit with coupled coils
As with any examples like this ONLY the leakage inductances produce an output impedance. For instance, if coupling is perfect (k=1) then the output terminals appear as a perfect voltage source with zero impedance.

Start with primary leakage inductance - this is \$L_1(1-k)\$. This will be "transformed" (or referred) to the secondary side by the turns ratio squared or \${L_2/L_1}\$. The rest of the primary (the coupled part) no longer plays any role in producing an impedance in series with the output terminals.

The referred_to_secondary primary leakage inductance will add to the actual leakage inductance of the secondary \$L_2(1-k)\$.

So looking into the terminals 1 and 2 you will see an inductance of: -

\$L_2(1-k) + L_1(1-k)\cdot{L_2/L_1}\$.

This looks like it reduces to an inductance of \$2\cdot L_2(1-k)\$

The above assumes that "E" is a perfect voltage source i.e. has zero source impedance. 

For the open circuit output voltage you have to calculate what is seen on the coupled part of the primary. Basically E is attenuated by the potential divider formed by leakage and coupled inductance.

This is simply \$\dfrac{k L_1}{kL_1 + (1-k)L_1}\$ or \$k\$.

So, the open circuit output voltage is turns ratio x k x E where turns ratio is \$\sqrt{L_2/L_1}\$.