Impulse response of a 3x3 PSF - how to find analytical expression for fourier transform of a 3x3 matrix?
You can use the 2D DFT formula

$$H(\omega_1,\omega_2) = \sum_{n_1} \sum_{n_2} h[n_1,n_2] e^{-j(\omega_1 n_1 + \omega_2 n_2)} $$

and simply the trigonometric algebra to get a closed form analytic expression for the 2D-DTFT. However, as @LaurentDuval has already mentioned, your 3x3 kernel is separable and one set of 1D filters is this

$$f[n_1]  = [\frac{1}{4} , \frac{1}{2}, \frac{1}{4}]^T $$
$$g[n_2]  = [\frac{1}{2} , 1 , \frac{1}{2}] $$

Then from the Fourier transform we know that

$$ h[n_1,n_2] = f[n_1]g[n_2] \implies H(\omega_1,\omega_2) = F(\omega_1) G(\omega_2) $$

Analytic expressions for 1D-DTFT's can be obtained from
$$H(\omega) = \sum_{n}h[n] e^{-j \omega n} $$

Assuming 3x3 kernel has zero phase, then we see that :
$$F(\omega_1) = \frac{1}{4} e^{j \omega_1 } + \frac{1}{2} e^{j 0 } + \frac{1}{4} e^{-j \omega_1 } = 0.5 \cos(\omega_1) + 0.5 $$

similarly 
$$G(\omega_2) = \frac{1}{2} e^{j \omega_2 } + 1 e^{j 0 } + \frac{1}{2} e^{-j \omega_2 } = \cos(\omega_2) + 1 $$

Hence:
$$H(\omega_1,\omega_2) = 0.5 \left( \cos(\omega_1) + 1 \right)(\cos(\omega_2) + 1) $$