For (2): let $G=\mathbb Z$, so $\phi$ is generated by a single homeomorphism of $X$. Let $X=\{ z\in\mathbb C : |z|\leq 1\}$ and define $\phi(re^{i\theta}) = r^2 e^{i\theta}$. Then $0$ and all points on the circle are fixed; the orbit of any other point $re^{i\theta}$ has accumulation points $0$ and $e^{i\theta}$. Let $f\in C(X)$ be invariant; translate so $f(0)=0$. Then $f(re^{i\theta}) = \lim_n f(r^{2n}e^{i\theta})=0$ for all $r<1$, so by continuity $f=0$. Thus the action $\alpha$ is "ergodic", but $\phi$ has non-trivial invariant open and closed sets.

Given there are no other answers, let me say something about (1). We need only check continuity at the identity of $G$. Let $(g_i)$ be a net converging to $e_G$. That $\alpha_{g_i}\rightarrow I$ means that for each $f\in C(X)$, we have $\|f\circ\phi_{g_i^{-1}} - f\|_\infty \rightarrow 0$. That $\phi_{g_i^{-1}} \rightarrow I$ means that whenever $K\subseteq X, U\subseteq X$ are compact (=closed) and open, respectively, with $K\subseteq U$, we have that $\phi_{g_i^{-1}}(K) \subseteq U$ for large $i$. (This is a basic open set about $I$ in the compact-open topology). To show that $\alpha$ continuous implies that $\phi$ is continuous, you can use a simple Urysohn's lemma argument (to find a function $1$ on $K$ and $0$ off $U$).

Conversely, suppose $\phi$ is continuous, and let $f\in C(X)$. For $\epsilon>0$ we can cover $f(X)\subseteq\mathbb C$ by a finite number of closed discs of radius $\epsilon/2$, say $(L_k)$. Then cover each $L_k$ by an open disc of slightly larger radius, say $V_k$. Then $K_k=F^{-1}(L_k)$ is closed in $X$, and $U_k=f^{-1}(V_k)$ is open, and contains $K_k$. So for $i$ large, by continuity of $\phi$, we have that $\phi_{g_i^{-1}}(K_k) \subseteq U_k$. Thus $f(x) \in L_k \implies x\in K_k \implies \phi_{g_i^{-1}}(x) \in U_k \implies \alpha_{g_i}(f)(x) \in V_k$. Hence $\|f - \alpha_{g_i}(f)\|_\infty$ is small. So $\alpha$ is continuous.

I think a good reference for the locally compact case is Dana Williams's book "Crossed Products of $C*$-Algebras". It nicely dots all the is and crossed all the ts.

For (2): let $G=\mathbb Z$, so $\phi$ is generated by a single homeomorphism of $X$. Let $X=\{ z\in\mathbb C : |z|\leq 1\}$ and define $\phi(re^{i\theta}) = r^2 e^{i\theta}$. Then $0$ and all points on the circle are fixed; the orbit of any other point $re^{i\theta}$ has accumulation points $0$ and $e^{i\theta}$. Let $f\in C(X)$ be invariant; translate so $f(0)=0$. Then $f(re^{i\theta}) = \lim_n f(r^{2n}e^{i\theta})=0$ for all $r<1$, so by continuity $f=0$. Thus the action $\alpha$ is "ergodic", but $\phi$ has non-trivial invariant open and closed sets.

Edit: An easier example has $X=[0,1]$ and $\phi(s)=s^2$. Then $\{0\}, \{1\}$ are non-trivial fixed closed sets, and $(0,1)$ is an invariant open set; but again $\alpha$ only leaves the constant functions invariant.