In the direction of question 1: if $X$ is compact Hausdorff and $Y$ has the homotopy type of a CW complex, then two maps $X\to Y$ are C-homotopic if and only if they are homotopic. In particular, C-homotopy groups agree with usual homotopy groups for CW complexes.
Edit: As Harry pointed out, the definition of homotopy groups uses relative homotopy, so an equivalence of C-homotopy and homotopy does not imply C-homotopy groups agree with homotopy groups.
For $X$, $Y$ arbitrary, let $\mathcal{C}(X,Y)$ be the space of continuous functions from $X$ to $Y$, equipped with the compact-open topology. If $X$ is locally compact Hausdorff, then for every $Z$ there is a bijection
$$
\{\text{continuous maps }Z\to\mathcal{C}(X,Y)\}\leftrightarrow\{\text{continuous maps }X\times Z\to Y\}.
$$
This implies that for $X$ locally compact Hausdorff, C-homotopy agrees with homotopy for maps $X\to Y$ if and only if every connected component of $\mathcal{C}(X,Y)$ is path-connected.
When $X$ is compact Hausdorff and $Y$ has the homotopy type of a CW complex, it is a theorem of Milnor that $\mathcal{C}(X,Y)$ has the homotopy type of a CW complex. It follows that the components of $\mathcal{C}(X,Y)$ are path-connected (this is true for CW complexes because they are locally path-connected, and a homotopy equivalence induces a bijection both on components and on path-components).
That said, connected components being path-connected is a much weaker condition than having the homotopy type of a CW complex, so probably there are much weaker hypotheses on $X$ and $Y$ under which C-homotopy agrees with homotopy.
Let $S^2$ be the unit sphere in $\mathbb R^3$, $S^2_+=\{(x,y,z)\in S^2\mid z\geq0\}$ the upper hemisphere, $S^2_-$ the lower hemisphere and $B^2$ the closed unit disk in $\mathbb R^2$. Define $p:S^2\to B^2$ by $p(x,y,z)=(x,y)$. Then the restrictions of $p_+:S^2_+\to B^2$ and $p_-:S^2_-\to B^2$ of $p$ are homeomorphisms. Define a copy of the topologist's sine wave in $B^2$ by $$W=\overline{\{(x,y)\in B^2\mid x\neq 0\text{ and } y = \tfrac12\sin\tfrac{\pi}x\}}.$$ Then, $K=p_+^{-1}(W)\cup p_-^{-1}(W)$ is a closed subset of $S^2$ with the following properties:
- $S^2\setminus K$ has two components $U_+$ and $U_-$ and the antipode of each point in $U_+$ lies in $U_-$ and vice versa.
- Antipodes of points in $K$ lie in $K$, but there is no path in $K$ between any such pair of antipodes.
Now consider $f:S^2\to\mathbb R^2$ defined by $$f(x,y,z)=(0,\pm d((x,y,z),K)),$$ where $d$ is e.g. the Euclidean metric on the sphere and the sign is $+$ for $(x,y,z)\in U_+$ and $-$ for $(x,y,z)\in U_-$.
This function $f$ has all the desired properties: it is continuous and the set of points $u=(x,y,z)$ such that $f(u)-f(-u)\in\mathbb R\times\{0\}$ is precisely $K$, because for any $u\in S^2\setminus K$, the second components of $f(u)$ and $f(-u)$ have different signs.
Best Answer
No, the harmonic archipelago (An illustration is on pg 7 of W. A. Bogley and A. J. Sieradski, Universal path spaces, preprint) is a locally path connected subspace of $\mathbb{R}^{3}$ and has uncountable fundamental group but every connected cover is trivial.
This is part of a more general phenomenon. Let $\pi_{1}^{top}(X)$ be the fundamental group of a space $X$ with the quotient topology of the loop space $\Omega(X)$ with the compact-open topology (sometimes called the "topological fundamental group"). This is a quasitopological group (in that inversion is continuous and multiplication is continuous in each variable) but is not always a topological group. If $p:X\rightarrow Y$ is a covering map the induced homomorphism $p_{\ast}:\pi_{1}^{top}(X)\rightarrow \pi_{1}^{top}(Y)$ is an open embedding of quasitopological groups (i.e. $\pi_{1}^{top}(X)$ embeds as an open subgroup). One consequence of this is that if $\pi_{1}^{top}(X)$ has the indiscrete topology, then either $\pi_{1}(X)=1$ or every connected covering of $X$ is trivial. There are lots of examples of spaces where this occurs other than the harmonic archipelago.