a)
Your start equation is missing a factor 6, which is the value of the resistor.
The frequency response would then be:
$$
G(S)=\frac{Y(S)}{U(S)}=\frac{3.(S+2)}{2.S+1}
$$
b)When the the frequency is zero the cap is an open circuit and the resulting gain should be 6, which is the value of the the resisor is parallel with the source.
When the frequency approaches infinity, the cap is a short circuit and the gain should be the parallel combination of the two resistor. That would be 12/8, you will reach the same value after calculating the limit with S-->infinity of the equation above.
Solution is all but trivial.
A rather good starting point would be doing the following assumptions valid for a resistive load:
1)Each cycle will start with C1 fully discharged and open TRIAC U2.
2)Load resistance is much lower than R3+RV1, hence you will have full half sine mains voltage across TRIAC.
3)DIAC is open circuit untill its breakover voltage VBO (approx 30V) is reached.
4)Now we have to write the transient of C1 being fed with sine voltage via R3+RV1.
5)When vc(t)=VBO TRIAC is fired, capacitor is discharged and your load is being fed.
So KVL to the source, R, C mesh would be
$$V_\text{max}\sin\omega t=v_\text{C}(t)+RC\,\frac{\text{d}\,v_\text{C}(t)}{\text{dt}}$$
the usual first order ODE to be solved in \$v_\text{C}(0)=0\$ boundary (or more generally some initial voltage as per Spehro's comment).
This is known to have solution sum of its general \$\breve{v}_\text{C}(t)=A\,\text{e}^{-t/\tau}\quad\tau=RC\$
and particular one \$\hat{v}_\text{C}(t)=\frac{V_\text{max}}{\sqrt{1+\omega^2\tau^2}}\sin\left(\omega t- \arctan(\omega\, \tau)\,\right)\$
Combining them in the above constraint and applying a little trigo gives
$$v_\text{C}(t)=\frac{V_\text{max}}{\sqrt{1+\omega^2\tau^2}}\sin\left(\omega t- \arctan(\omega\, \tau)\,\right)+\frac{V_\text{max}\,\omega\, \tau}{1+\omega^2\tau^2}\text{e}^{-t/\tau}$$
which equated to DIAC break over would give TRIAC on time.
$$\frac{V_\text{BO}}{V_\text{max}}=\frac{\sin(\omega\, t_\text{on}- \arctan(\omega\, \tau)\,)}{\sqrt{1+\omega^2\tau^2}}+\frac{\omega\,\tau\,\text{e}^{-t_\text{on}/\tau}}{1+\omega^2\tau^2}$$
What we really understand from the above is that's indeed job for a numeric solutor.
Edit: one sign fixed upon @Delfin suggestion.
Best Answer
Yes, why not?
Yes. It's obvious that the pulse response is a pure harmonic function because it is a resonance circuit without resistance.
Let the pulse be a voltage dirac pulse of value \$1 V\$ at \$t=0\$, and the output signal the voltage over the capacitor. Then it's clear that at \$t=0\$, \$U_C=1 V\$ and that the impulse response is a harmonic voltage starting at \$U_C = 1 V\$ with an angular frequency \$\omega_0 = \frac{1}{\sqrt{L C}}\$ (IIRC).
Then \$h(t)=cos(\omega_0 t)\$
Why bothering you with that?
Well, because \$\int_\infty^\infty cos(\omega_0 t) e^{-j \omega t}dt = \frac{1}{2}[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]\$.
I guess that integrating from 0 to \$\infty\$ gives only the \$\delta(\omega-\omega_0)\$ term.