The fundamental group(s) of $\Bbb Q$ and of $\Bbb R$.

Any continuous map $\Bbb S^1\to \Bbb Q$ is constant, therefore the fundamental group of $ \Bbb Q$ is trivial no matter what the base point is.

Any continuous map $f:\Bbb S^1\to \Bbb R$ can be continuously shrunk to a constant map via $F(x,t)=(1-t)f(x)$, where $t=0$ yields $f$ and $t=1$ yields a constant map. Therefore again the fundamental group of $ \Bbb R$ is trivial no matter what the base point is.

Hint: $\pi(X,x_0) \cong \pi(X',x_0)$ where $X'\subset X$ stands for the path-component of $X$ containing $x_0$