The correct value of $\omega$ is not $\frac{\sqrt{7}}{2} \approx 1.32$. It's $\sqrt{\frac{3}{2}} \approx 1.22$. We are looking for a near resonance effect, but you can't actually have resonance when the harmonic oscillator is damped. This makes the analysis is a little different.
A reference is the section on sinusoidal forcing in the Wikipedia page on the harmonic oscillator. From the formulas there you can see that the relationship between $A$ and $\omega$ is given by
$$A = \frac{1}{\sqrt{\omega^2 + (2 - \omega^2)^2}}.$$
Solving $\frac{dA}{d \omega} = 0$ yields $\omega = \sqrt{\frac{3}{2}}$.
Update: Here's the derivation.
The trick is to generalize, and solve the differential equation $x'' + x' + 2x = e^{i \omega t}$. The resulting solution will have a real part and an imaginary part. Since $e^{i \omega t} = \cos \omega t + i \sin \omega t$, you actually want the imaginary part of the solution.
As the driving force is an exponential, we know that the particular solution must be of the form $x_p(t) = c e^{i \omega t}$. Subbing that into the differential equation produces the auxiliary equation $-c \omega^2 + i c\omega + 2c = 1$. Solving that for $c$ yields $$c = \frac{1}{a + i b} = \frac{a - i b}{a^2 + b^2},$$
where $a = 2 - \omega^2$ and $b = \omega$. Thus the particular solution to the complex differential equation is
$$x_p(t) = \frac{a - i b}{a^2 + b^2} (\cos \omega t + i \sin \omega t),$$ of which the imaginary part is $$-\frac{b}{a^2 + b^2} \cos \omega t + \frac{a}{a^2+b^2} \sin \omega t.$$ Since $A$ is just the magnitude of this solution (you're doing a rotation to the vertical axis when converting to $A \sin (\omega t + \phi)$), we get $$A = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{\sqrt{\omega^2 + (2 - \omega^2)^2}}.$$
