In Fourier analysis we are trying to see how sinusoids over all frequencies are scaled and added to define a time signal $x(t)$. Usually we have a time signal, do Fourier analysis, and we get a new function $X(f)$ that shows how each sinusoid at a frequency $f$ contributes to creating the time signal. This is called the spectrum of $x(t)$.
When first learning this theory, there is something that is often not stressed enough about Fourier analysis, and the Fourier transform specifically: it assumes that the sinusoids used for the analysis are complex. That is of the form
$$e^{j2{\pi}ft}$$
Any other details about signals being real or complex and how they show up on the spectrum stems from this assumption.
You can see this from the definition of the Fourier transform where the input signal is multiplied by the complex sinusoids over all frequencies
$$X(f) = \int_{-\infty}^{\infty}x(t)e^{-j2{\pi}ft}dt$$
Let's take the example of $x(t) = e^{-j2{\pi}f_0t}$. Taking the Fourier transform we get
$$X(f) = \int_{-\infty}^{\infty}e^{-j2{\pi}f_0t}e^{-j2{\pi}ft}dt = \delta(f-f_0)$$
This should be intuitively satisfying. Since our time signal is a complex sinusoid at some frequency $f_0$, our spectrum results in a single "spike" that only exists when $f = f_0$. In other words, the result tells us that only the complex sinusoid of the frequency $f_0$ contributes to the signal.
Now let's take the example where our time signal is now the real sinusoid $x(t) = cos({2{\pi}f_0t})$. A transform table will tell you that the Fourier transform is
$$X(f) = \frac{1}{2}[\delta(f - f_0) + \delta(f + f_0)]$$
Why do the two "spikes" show up? Why is it that this result looks like the one above, but now includes a mirrored version on the other side of the spectrum?
Recall the result
$$cos(x) = \frac{e^{-jx} + e^{jx}}{2}$$
A real sinusoid can be seen as the sum of two complex sinusoids.
Using our signal $x(t)$ we get
$$cos(2{\pi}f_0t) = \frac{e^{-j2{\pi}f_0t} + e^{j2{\pi}f_0t}}{2}$$
Treating each term separately and taking the Fourier transform, you will yield the sum of the two Dirac deltas.
Applying this to your example, which assumes the angular frequency $\omega$:
$$x(t) = cos(50t) + 2cos(70t)$$
$$ = \frac{e^{-j50t} + e^{j50t}}{2} + ({e^{-j70t} + e^{j70t}})$$
Following the same logic as above and applying the angular-frequency Fourier transform you should see why you get Dirac deltas at $\omega = -70, -50, 50, 70$.
Sampling
Using the sampling theorem, it states that when sampling at a rate of $f_s$, the resulting discrete signal now has a spectrum that contains copies of the original spectrum centered around all multiples of $f_s$. In your example, you are actually under-sampling and therefore aliasing.
Your sampling period results in a angular-frequency of $\omega_s = 1/T_s \approx 19 \space rads/s$. When the frequency band is referenced, the problem is asking you to only consider the chunk of the spectrum between $\omega = -55$ and $\omega = 55$.
Since you will be aliased, copies from the $cos(70t)$ term will actually appear at $\omega = -50$ and $\omega = 50$. This is the same result as if you sampled the original signal correctly, that is, you met the Nyquist criteria.
You should really try and simulate this. It will accelerate mastering and becoming comfortable with these concepts.
