Could you explain to me how they solved this problem using the charatheristics method (transport equation)? I don't understand how to derive the charateristics, do I integrate? Do I differentiate? I have no idea.
In their solution, I don't understand how they derived those fractions (or anything really).
Note that if you take a curve $(t,x(t))$ with the property that $$ x'(t)=x(t)^2t $$ Then by the chain rule $$ \frac{d}{dt}u(t,x(t))=0 $$ for any $u$ satisfying your pde. In other words, any potential solution will be constant along curves satisfying the ode, so given initial condition for our curve $(0,x(0))=(0,x_0)$, we have, thanks to the initial data given, $$ u(t,x(t))=\cos(x_0) $$ This observation suggest a way to build solutions (in fact provably the unique solution, using ode theory of existence and uniqueness).
So, what is $x(t)$? solving the ode via an easy separation of variables, we find $$ \frac{1}{x}=\frac{C-t^2}{2}\implies x=\frac{2}{C-t^2}\ \implies x=\frac{2}{\frac{2}{x_0}-t^2} $$ where we take $x_0$ an arbitrary (nonzero) initial value on the $x$ axis in the $(t,x)$ plane on which $u$ is defined. This is a good choice since we know what our solution ought to be doing at $t=0$, it is just $\cos(x_0)$ by your initial condition. Since again, our solution is constant along these projected characteristics, it will stay $\cos(x_0)$.
Since we don't want to solve uncountably many of these odes (one for each $x_0$), it would be good to solve for $x_0$ in terms of $x$ and $t$. This is readily done as $$ \frac{2}{x_0}=\frac{2}{x}+t^2\implies x_0=\frac{x}{1+\frac{1}{2}xt^2} $$ and your solution is $$ u(t,x)=\cos\left(\frac{x}{1+\frac{1}{2}xt^2} \right) $$ Where we assumed $x_0\ne 0$ in order to divide. However, if $x_0$ is zero, then we have that the pde is just $1=\cos(0)$ along the characteristics.