Why does the follsolution of the ODE $x'=(t.cos(t)+sin(t),t^2.cos(t)+2t.sin(t))$, $x(0)=(0,0)$ doesn't contradict Picard? I have found the following solution:
$\varphi(t)=(t.sin(t),t^2.sin(t))$, but $\varphi(0)=\varphi(2\pi)$, while $\varphi'(0)$ and $\varphi'(2\pi)$ are linearly independent. My professor said that this answer is correct, but he told me to think a bit more why it doesn't contradict Picard's theorem. But I can't find a good reason.
 A: The way you seem to be thinking about Picard's theorem is the way one should think about it for autonomous systems. In this case, if $y$ solves your ODE and $y(a)=y(b)$, then $y'(a)=y'(b)$ is one of several consequences of Picard's theorem. This is what you seem to be expecting.
But your equation, which is (from the comments):
$$\frac{dx_1}{dt}=t \sin(t) + \cos(t) \\
\frac{dx_2}{dt}=t^2 \cos(t)+2t\sin(t)$$
is not autonomous. In fact it doesn't even have explicit dependence on $x$ at all. However, you can always* make an ODE autonomous by making time a dependent variable, and introducing a dummy independent variable. In your case this looks like:
$$\frac{dx_1}{ds}=t \sin(t)+\cos(t) \\
\frac{dx_2}{ds}=t^2 \cos(t)+2t\sin(t) \\
\frac{dt}{ds}=1.$$
In this formulation, your equation is autonomous, but the initial data now consist of three numbers, $(x_1(t_0),x_2(t_0),t_0)$. In your case the first two agree but the third does not, so there is no contradiction to Picard's theorem.
* This is not quite true: Picard's theorem for $y'=f(t,y)$ only requires $f$ to be continuous in $t$ and Lipschitz continuous in $y$. "Autonomization" would require $f$ to also be Lipschitz continuous in $t$. But there is no problem in your case, at least on a finite time interval.
