Electronic – How to solve for the Norton Current of this circuit

This is just a matter of references*.

When you start to tackle the problem you short circuit your port and draw a reference for the Norton current. It's just like saying "this is what positive Norton current is to me".

After solving the circuit you get a result that is negative? No problem: it just means that the current is actually flowing in the opposite way, i.e. out of node 1. Your answer and your book's are just the same: the book says that output current is positive, and gets a positive current that is then going out, while you say that input current is positive, and get a negative current that is then going out.

Note that an answer giving only the current would be incomplete: if I told you that the current flowing in the controlled source is 1A you should ask me: is it going from plus to minus or vice versa?

You can think of an analogy in kinematics: the first thing you do before solving a problem is choosing a reference system*. If you don't choose it and tell your teacher that the ball speed is 10m/s he should yell at you.

*I believe that's the correct English word, please comment if it's not.

We're using superposition to establish the short circuit curent. When ignoring \$i_1\$ we can calculate the impedance between terminals A and B:

$$ Z_{AB} = j \omega L_2 + \frac{1}{j \omega C + 1/R} $$

leading to a current of $$ i_{u1} = \frac{u_1}{Z_{AB}} = \frac{u_1}{j \omega L_2 + \frac{1}{j \omega C + 1/R}}$$

This is the first partial current. Now, if we consider \$i_1\$ and replace \$u_1\$ with a short circuit, \$i_1\$ is connected directly between terminals A and B. Thus:

$$ i_{short} = i_1 + i_{u1} $$

So you are half correct. The short circuit current consists of a part \$i_1\$ and a part \$i_{u1}\$ contributed by the voltage source.