1) 7884 seconds (that's over two hours).  The initial voltage on the capacitor is 6 volts, and the RC time constant is 4400 seconds.  We want to know the time that the voltage has decayed to 1 volt, so we set the transient expression  vc = 6 exp (-t / 4400)  with vc  = 1, and solve for t.

2) v(0+) = 2.5 volts (-> when the switch has been closed for a long time, the capacitor charges up and its current becomes zero, so the 5V source is split between the two 10k resistors).

v(inf) = 0 volts (-> after the switch opens the capacitor voltage decays, eventually reaching zero)

time constant = RC = 10k x 1 uF = 10 milliseconds (-> after switch opens, the only resistance seen by the capacitor is the right 10k resistor)

3) Figure b is the correct one.  The inductor current is zero at t =0-, and cannot change instantaneously at t =0 when the switch closes.  The inductor current then must increase exponentially and asymptotically toward its final value, which would be V/R.

4) Vth is 13.33 at 0 deg.  The Thevenin voltage is the open circuit voltage "looking in" at the indicated port.  Since it is an open circuit, there is no current in the inductor and therefore no voltage across the inductor, so the open circuit voltage is the same as the voltage across the 20 ohm resistor.

Zth is 6.67 +j10 ohms.  Impedance seen looking into the port with the independent source turned off, which will be the inductor impedance in series with 10 and 20 ohms in parallel with each other.  Inductor impedance is j w L, and w = 100 rad/sec, so inductor impedance is j10 .

Increasing the frequency will cause the impedance to increase, because j w L will increase.

5) The second expression is the correct one.  Notice that the three impedances are in parallel with each other, so the equivalent impedance is  1 / ( 1/Zc) + (1/ZL)  +  (1/R)  ).  The algebra can be tricky, but manipulate the denominator to get the expression in the indicated form

6) Looking for the resistor current.  Note that with the switch closed for t <0, the inductor current will be zero.  When the switch opens the inductor current cannot change instantaneously, so iL(0+) = 0, and therefore the 2A source current must all go through the resistor.  Thus, iR(0) = 2.  Then after a long time the inductor will be conducting current and eventually its voltage will go to zero so that the inductor conducts the entire 2A as t goes to infinity.  Thus, IR(inf) = 0.  The L/R time constant is 0.2 seconds.

So:

iR(t) = 2 exp (-t / 0.2)  amps for t>0