In which condition the current through the inductor approach INFINITY
and how can you explain that physically?
Yes, it is absolutely true - the current in the inductor can become infinite and the current in the capacitor can also become infinite however, because these two currents are anti-phase (one lags the sinewave source by 90deg and one leads the sinewave source by 90deg), the net current taken from an AC voltage source is zero other than what is consumed in the parallel resistor.
You are using a current source and the resultant voltage will be I*R. I'm also presuming you meant R to be a positive resistance value not as shown in your diagram although that won't alter things at all - if you meant R to be negative then the voltage developed across the resistor will be anti-phase to the current through it.
The actual current through the inductor is the voltage across it (as developed by the resistor and current source). Ditto the capacitor.
To find the Thevenin equivalent resistance you need to turn off the independent current and voltage sources. A current source that has been turned off has 0A and is therefore an open circuit. A voltage source that has been turned off has 0V and is therefore a short circuit.
For this circuit, therefore, \$R_1 + R_2 = 20\Omega\$ is in parallel with \$R_3 = 5\Omega\$ (you can ignore \$I_1 = 0\$, and the lower ends of \$R_1\$ and \$R_3\$ are shorted since \$V_1 = 0\$).
\$(R_1 + R_2)||R_{3} = 20\Omega ||5\Omega = 4\Omega\$
This \$4\Omega\$ resistance is in series with \$R_5 = 1\Omega\$, giving \$5\Omega\$ resistance. This \$5\Omega\$ resistance is in parallel with \$R_4 = 10\Omega\$ so the Thevenin resistance across \$V_{o}\$ is \$5\Omega || 10\Omega\ = 3.33\Omega\$.
It's a similar procedure to find \$V_{TH}\$ except that the sources are left on. Combine the resistances where possible as I did for \$R_{TH}\$ to find the voltage at the node common to \$R_{2}\$, \$R_{3}\$, and \$R_{5}\$. Then \$R_{5}\$ and \$R_{4}\$ form a voltage divider which gives you the voltage \$V_{TH}\$ across \$R_{4}\$.
To calculate \$V_{TH}\$ use superposition: calculate \$V_{TH}\$ with the current source turned off (\$I_1 = 0\$), then calculate \$V_{TH}\$ with the voltage source turned off (\$V_1 = 0\$), and add the two results to find \$V_{TH}\$ as a result of both sources.
With \$I_1 = 0\$, \$R_1 + R_2\$ is in parallel with \$R_4+R_5\$. The voltage at the top node (call it \$V_{t1}\$) is calculated from a voltage divider formed by \$R_3\$ and \$(R_1 + R_2)||(R_4 + R_5)\$. Then looking back at the original circuit, you can calculate \$V_{TH1}\$ from the voltage divider of \$V_{t1}\$ formed by \$R_4\$ and \$R_5\$. \$V_{TH1}\$ is \$V_{TH}\$ due to \$V_1\$ only.
With \$V_1 = 0\$, \$R_4+R_5\$ is in parallel with \$R_3\$. \$(R_4+R_5)||R_3\$ is in series with \$R_2\$. \$((R_4+R_5)||R_3) + R_2\$ is in parallel with \$R_1\$ so use a current divider to find the current flowing into \$((R_4+R_5)||R_3) + R_2\$. This is the current flowing into \$R_2\$ in the original circuit, so use a current divider again to find the current flowing through \$R_3\$. This current multiplied by \$R_3\$ is \$V_{t2}\$, and you can calculate \$V_{TH2}\$ from the voltage divider of \$V_{t2}\$ formed by \$R_4\$ and \$R_5\$. \$V_{TH2}\$ is \$V_{TH}\$ due to \$I_1\$ only.
By superposition \$V_{TH} = V_{TH1} + V_{TH2}\$.
Best Answer
A general way to do this is to solve for the total complex impedance, then find the frequency that minimizes the impedance magnitude.
A more case-specific way is to notice that the resistor is in series with a series L-C. Consider the impedance of a series L-C. It is infinite at 0 frequency due to the capacitor. It is infinite at infinite frequency due to the inductor. It is actually 0 at the resonant frequency, and goes up either side of it.
Therefore use the standard formula for finding the resonant frequency of a L-C system. At that frequency, all the voltage applied by V1 will appear across R1, and R1 will dissipate V12/R1.
Actually cranking thru the numbers is your job.