I assume that exercise is meant to demonstrate the difference in periodicity between functions from a continuous domain (like the real or complex numbers) and from a discrete domain (like the natural numbers/integers).
1) This function is periodic when discretized as it happened here. Try to prove that $$x[n+4]=x[n]$$ by taking into account that the sine and cosine functions are periodic with any integer multiple of $2\pi$.
2) This function is not periodic. $x[0]=\cos(\pi)=0$. If there was some $N>0$ with $x[n+N]=x[n]$ for all $n$, then it must be true that $x[N]=0$. The cosine function is $0$ only for integer multiples of $\pi$ (more accurately, it is $0$ for exactly the odd integer multiples of $\pi$, but the less accurate version is enough). $x[N]=0$ implies there is some integer $k$ that
$$\sqrt{2}\pi N +\pi = k\pi$$
Dividing by $\pi$ results in
$$\sqrt{2}N +1 = k,$$
which can be transformed into
$$\sqrt{2}=\frac{k-1}{N}$$
since $N > 0$. But this is impossible since $\sqrt{2}$ is an irrational number. $\blacksquare$
As you correctly wrote the fundamental period of the $\mathbb R \rightarrow \mathbb R$ function $f(t) = \cos(\sqrt{2}\pi t + \pi)$ is $\sqrt{2}$. That means for question 2) the underlying $\mathbb R \rightarrow \mathbb R$ function is periodic, while the discrete version is not.
It is the other way around for question 1). Here the discretized version is periodic, while the $\mathbb R \rightarrow \mathbb C$ function $f(t)=e^{j\frac{3}{4}\pi t^2}$ is not. There are many ways to visualize that. Let's just look at the real part of $f: Re(f(t)) = \cos(\frac{3}{4}\pi t^2)$.
The derivative of it is
$$ (Re(f(t)))' = -\frac{3}{2}\pi t\sin(\frac{3}{4}\pi t^2)$$
which is unbounded when $t$ becomes large. But the derivative of a periodic function is itself periodic (say with period $p$), and the value range over all reals is the same as the value range over the interval $[0,p]$. Since $(Re(f(t)))'$ is continuous, its value range over a finite closed interval must be bounded.
I am not very sure where you are getting at and what you mean by "frenquency going to $0$". At least in your example the frenquency $\omega_0$ is always a constant.
As I understand, the integral
$$\frac{1}{a-b}\int_a^bf(x)dx$$
is the average of $f(x)$ over the interval $[a,b]$.
If $f(x)$ has period $T$, then one can prove that
$$\frac{1}{T}\int_a^{a+T}f(x)dx\quad(*)$$
is a constant regardless of the choice of $a$. One might say in this case that $f(x)$ has a "well-defined" average.
If, unfortunately, $f(x)$ is not periodic but we still want to talk about the "average" of $f(x)$, we consider its average over some finite interval $[-T/2,T/2]$
$$\frac{1}{T}\int_{-T/2}^{T/2}f(x)dx$$
and take the limit $T\to\infty$ and call the limit (if it exists)
$$\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}f(x)dx$$
the average of $f(x)$ over $\mathbb{R}$.
It can also be shown that the second average is consistent with the first in the case where $f(x)$ is periodic. My opinion is that the limit definition of the average of a function is simply a generalization to deal with not strictly periodic functions. In your case, power is the average of energy over time, so you may understand the limit as the average of the energy of a signal over a very long time.
But still, I don't see how this relates to any change in frenquency.
Best Answer
First thing to note is that ${\rm rect}(t)^2$, at least for the purposes of integration, is equal to ${\rm rect}(t)$. This is trivially true for all $|t|\neq1/2$, and the value at the single points $t=\pm1/2$ has no effect on the value of the integral, as long as no dirac functions are involved. So our integral becomes:
$$P=\frac{1}{T_0}\int_{T_0}(2{\rm rect}(\sin(2\pi t)))^2\ {\rm d}t= \frac{1}{T_0}\int_{T_0}4{\rm rect}(\sin(2\pi t))\ {\rm d}t$$
Now, the way I'm familiar with handling ${\rm rect}$ and unit step functions in integrals, is to change the bounds. But first, we need to find which values of $t$ over one period satisfy $|\sin(2\pi t)|<1/2$. First note that for one period of the function, it we get two intervals of $t$ that satisfy the inequality (one with $\sin(2\pi t)$ increasing, one decreasing). The endpoints of these intervals can be found by solving the equality:
$$\sin(2\pi t)=\frac{1}{2}\quad\Rightarrow\quad t=\frac{\sin^{-1}(1/2)}{2\pi}=\frac{\pi/6}{2\pi}=\frac{1}{12}$$ $$\sin(2\pi t)=-\frac{1}{2}\quad\Rightarrow\quad t=\frac{\sin^{-1}(-1/2)}{2\pi}=\frac{-\pi/6}{2\pi}=-\frac{1}{12}$$
Remembering that $y=\pi-\sin^{-1}(x)$ is also a solution to $\sin(y)=x$, we get the endpoints of the other interval:
$$\sin(2\pi t)=\frac{1}{2}\quad\Rightarrow\quad t=\frac{\pi-\sin^{-1}(1/2)}{2\pi}=\frac{5\pi/6}{2\pi}=\frac{5}{12}$$ $$\sin(2\pi t)=-\frac{1}{2}\quad\Rightarrow\quad t=\frac{\pi-\sin^{-1}(-1/2)}{2\pi}=\frac{7\pi/6}{2\pi}=\frac{7}{12}$$
Plugging in $T_0=1$ and choosing our period to be $[-1/4,3/4)$, we get:
$$P= \frac{1}{T_0}\int_{T_0}4{\rm rect}(\sin(2\pi t))\ {\rm d}t=\frac{1}{1}\left(\int_{-1/12}^{1/12}4\ {\rm d}t+\int_{5/12}^{7/12}4\ {\rm d}t\right)\\ =\left(\frac{1}{12}-\frac{-1}{12}\right)4+\left(\frac{7}{12}-\frac{5}{12}\right)4\\ =\left(\frac{2}{12}+\frac{2}{12}\right)4\\ =\frac{4}{3}$$