Inverse Fourier transform from plots
In this case, it is quite simple. In general, it may be more difficult. I'm not sure what you are getting at with $H(j \omega)$ since you don't define it anywhere, but you are trying to get the analytical time domain expression for the signal from the plots. This is possible by noticing that
$$X_1(j \omega) = |X_1(j \omega)| \exp\left( j \angle X_1(j \omega)\right).$$ 
In this case, due to the plots' simplicity, we can generate a formula by reading directly from the axes. In this case,
$$X_1(j \omega) = 3 j \omega \left[ u(\omega + 3 \pi) - u(\omega - 3 \pi) \right],$$
where $u(t)$ is the heaviside step function.

Now, we can calculate the Inverse Fourier Transform with the formula,
$$ x_1(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X_1(j \omega) \exp\left(j \omega t \right) d \omega. $$
Plugging in our $X_1(j \omega)$ and simplifying leads us to the expression 
$$ x_1(t) = \frac{1}{2 \pi} \int_{-3 \pi}^{3 \pi} 3 j \omega \exp\left(j \omega t \right) d \omega.$$

I assume that you can evaluate this integral. If you need more assistance, let me know.