Find the power at the output of a random input + noise?
Point A is quite straightforward.

To achieve \$\text{SNR}=1000\equiv30\text{dB}\$ you will need the output signal power to be 1000 times higher the output noise power. The input noise is white, meaning its PSD is flat. The value is given, i.e. \$10^{-3}\frac{W}{Hz}\$. (They should really provide measurement units in my opinion.) \$H(\omega)\$ is an ideal low pass filter, the cutoff frequency is \$f_c=\frac{\omega_c}{2\pi}=159Hz\$. The output noise power is thus:
$$
P_n^{out}=\int_{-\infty}^{+\infty} S_n^{in}(f)\cdot|H{f}|^2df=2\int_0^{f_c}10^{-3}\cdot4df=
1.27W
$$
Yeah you've got a HUGE input noise. Finally:
$$
P_s^{out}=1000P_n^{out}=1.27kW
$$

Now to part B. You are given the input signal power spectral density. You can calculate the output powers (signal and noise) independently because \$S_{XX}(\omega)\$ is independent from \$M(t)\$. The output signal power is now:
$$
P_s^{out}=\int_{-\infty}^{+\infty}S_{XX}(\omega)|H{\omega}|^2d\omega=
2\int_0^{10^3}\frac{c}{1+\omega^2}\cdot4d\omega=8c\left[tan^{-1}(\omega)\right]_0^{10^3}\approx c\cdot 12.55W
$$

Finally:
$$
c = \frac{1.27kW}{12.55W}\approx 101
$$

Please note that since \$S_{XX}(\omega)\$ is a second order low pass function with the poles in \$\omega=1\$ you could have probably used some sort of approximation. In this case, for the signal, the value of the filter cutoff frequency does not matter much: the signal PSD is so low up there that if the cutoff frequency happened to be several times higher the signal power would have been remained the same. The noise power instead is flat, so the cutoff frequency matters a lot. I pinpointed this fact because you will be in this situation very often.