Why is $A\cos(2\pi f_ct)$ a non-stationary process?
A random process is a collection of random variables, one random variable for each time instant. It is best to write the
random process as
$$\{X(t)\colon -\infty &lt; t &lt; \infty\} \tag{1}$$
where the $\{$ and $\}$ indicate that a set (or collection) of
objects is being
defined, and the interior says that a typical member of this set is denoted
by $X(t)$ where $t$ is a real number. Think of $X$ as the family name of all the random variables in the set, and $t$ as the first name. Thus, $X(t_0)$ is just one of the random variables in this collection of random variables, the random variable for time $t_0$, while $X(t_1)$ is another random variable, the one corresponding to time $t_1$.  

The statement
$$X(t) = A\cos(2\pi f_ct)$$
is a loose way of writing the more formal definition $\{X(t) = A\cos(2\pi f_c t)\colon -\infty &lt; t &lt; \infty\}$. Thus, we have that
$$X(t_0) = A\cos(2\pi f_ct_0),\\
X(t_1) = A\cos(2\pi f_ct_1),\\
\cdots $$

Since $\cos(2\pi f_ct_0)$, $\cos(2\pi f_ct_1)$ etc are just constants,
what we immediately get is that each random variable $X(t_i)$ in
this random process is a uniformly distributed random variable. Huh?
How did we get that? Well, $A \sim U(0,1)$ is given, and hopefully
you understand enough about random variables to know (or to be able
to jump
to the conclusion) that $bA \sim U(0,b)$ for $b &gt;0$.  If $b &lt; 0$,
then $bA \sim U(b,0)$.

So, now you have enough information to decipher the crappy notation
in the solution given to you (which is not completely correct, anyway). The random variable $X(t_i)$ is given by
$$X(t_i) = A\cos(2\pi f_ct_i) = bA  
\begin{cases} \sim (U(0,b), &amp; b &gt; 0,\\
= 0, &amp; b = 0,\\\sim U(b,0), &amp;b &lt; 0,\end{cases}.$$
Thus, for $b &gt; 0$, we can write the density function of $X(t_i)$ as
$$f_{X(t_i)}(x) = \begin{cases} \displaystyle\frac 1b, &amp; 0 &lt; x &lt; b,\\
0, &amp;\text{otherwise},\end{cases}
= \begin{cases} \displaystyle \frac{1}{\cos(2\pi f_ct_i)}, &amp; 0 &lt; x &lt; \cos(2\pi f_ct_i),\\
0, &amp;\text{otherwise},\end{cases} \tag 2$$
where of course $t_i$ is such that $\cos(2\pi f_ct_i) = b$.

So, the above derivation of $(2)$ answers your first question. For the
second question re stationarity, the conditions that a random process
must satisfy in order for it to be called a stationary process are
quite onerous (see, for example, the first paragraph of
this answer of mine to
a different question for details) and this process fails to meet
even the simplest of the necessary conditions: that all the random
variables comprising the process have the same distribution function
(which also means having the same density function if the random
variables are continuous random variables). But, it is manifestly
obvious from $(2)$ that $X(t_i)$ and $X(t_j)$ have different
densities in general. That for certain specific choices of
$t_i$ and $t_j$, $X(t_i)$ and $X(t_j)$ have the same density (in fact
are the same random variable) is nice to know, but irrelevant. 
Not all random variables in the process can be said to
have the same
distribution, and this single inconvenient fact allows us to reject 
any overblown notions that we might be harboring 
that this process is a stationary process.

To avoid follow-on questions from readers, the process in question
is not wide-sense stationary either (hint: because the mean function
is time-varying).