Filtering Passband Signals using Complex Baseband Filtering
Since $h_I(t)$ is the real part of $h(t)$ you have

$$h_I(t)=\frac12[h(t)+h^*(t)]\tag{1}$$

where $h^*(t)$ is the complex conjugate of $h(t)$. For the imaginary part you have

$$h_Q(t)=\frac{1}{2j}[h(t)-h^*(t)]\tag{2}$$

Since the Fourier transform of $h^*(t)$ is $H^*(-f)$1, the Fourier transforms of (1) and (2) are

$$H_I(f)=\frac12[H(f)+H^*(-f)]\\
H_Q(f)=\frac{1}{2j}[H(f)-H^*(-f)]$$


1The fact that $H^*(-f)$ is the Fourier transform of $h^*(t)$ can be easily seen as follows:

With
$$H(f)=\int_{-\infty}^{\infty}h(t)e^{-j2\pi ft}dt$$
the Fourier transform of $h^*(t)$ is
$$\mathcal{F}\{h^*(t)\}=\int_{-\infty}^{\infty}h^*(t)e^{-j2\pi ft}dt=
\left[\int_{-\infty}^{\infty}h(t)e^{j2\pi ft}dt\right]^*=H^*(-f)$$