It's actually quite straightforward to find the Thevenin impedance of this circuit.

The equivalent impedance looking into the port ab is defined by:

$$Z_{ab} \equiv \frac{V_{ab}}{I_{ab}} = \frac{V_{ab}}{i_{\Delta}} = Z_{th}$$

But you can write by inspection a simple KVL equation for \$V_{ab}\$ in terms of \$i_{\Delta}\$:

$$V_{ab} = i_{\Delta}(-jX_{1uF} + 10k\Omega) + 200 i_{\Delta} \cdot 100 \Omega) = i_{\Delta}(-jX_{1uF} + 30k)\Omega$$

Generally speaking, to find the Thevenin equivalent of a circuit with only dependent source(s), you must be sure to "activate" the dependent source(s) with a test source.

This is what was done above. We solved for the voltage across the port due to a test current source, \$I_{ab}\$ which, in this case, equals the controlling variable thus making this problem particularly easy to solve.

A key thing to remember is that the current in an (idealised) inductor cannot change instantaniously but the voltage can (this is the opposite of the ideal capacitor where the voltage cannot change instantanously but the current can)

There are three stages to consider (I will assume all variables use si units).

Before t=0:

The voltage and current in the inductor are zero.

Between t=0 and t=10^-3

We can replace the voltage source and two resistors with a thevin equivilent circuit. This works out to a 24V supply with a 100 ohm series resistance. We now have a series LR circuit supplied by a 24V supply with zero current at t=0

We could solve the series LR circuit from first principles but I just went and looked it up on google and found http://www.electronics-tutorials.ws/inductor/lr-circuits.html . This tells me that the equation for this case is.

$$I=\frac{V}{R}(1-e^{-Rt/L})$$

Where I is the current in the indctor, L is the inductance and V and R are the 24V and 100 ohm we got from our thevinin calculation above.

After t=10^-3

The only components relavent now are the inductor and the 125 ohm resistor. The current in the inductor at the start of this phase will be the same as the current in the inductor at the end of the previous phase but it is now forced to go through the resistor, this will cause the voltage on the inductor to reverse and the energy stored in the inductor to discharge. I'll leave you to either work out and solve the differential equation or google for the equation needed in this case.