As $\gamma$ is generally used for curves, i will be using $\Gamma$ and $\Gamma'$ for the fuschian groups and $p$ and $p'$ the corresponding covering maps.
a) Let the two groups be conjugate by a function $h$ i.e $h\Gamma h^{-1}= \Gamma'$. Define $f(x)=h(y)$ for some $y\in p^{-1}(x)$. We will check that this is well defined. That is we need to check $p(y)=p(y')$ then $p'(h(y))=p'(h(y'))$. $p(y)=p(y')$ implies that $y'=g(y)$ for some $g\in \Gamma$. $h\circ g\circ h^{-1} = g'\in \Gamma'$ which implies that $h\circ g = g^{-1}\circ h$. $h(y')=h(g(y))=g'(h(y))$ thus $p'(h(y))=p'(h(y'))$.
This map is clearly a local isometry. An assumption that $f$ is nor injective(surjective) will imply a similar statement for $h$, a contradiction. Thus $f$ is bijective.
To prove part b) i will use the following standard results from covering space theory(c.f Hatcher: Algebraic topology)
1) For a fixed point $z_0\in p^{-1}(x_0)$ ($p$ is the covering map) the set $\Gamma$ corresponds one-one to $p^{-1}(x_0)$ via $g\in\Gamma\leftrightarrow g(z_0)\in p^{-1}(x_0)$.
2) For a fixed point $z_0\in p^{-1}(x_0)$ the group of deck transformations $\Gamma$ and $\pi_1(\mathbb{H}/\Gamma,x_0)$ are isomorphic. The isomorphism is: an element $\gamma\in \pi_1(\mathbb{H}/\Gamma,x_0)$ is sent to the deck transformation such that $z_0\mapsto \tilde{\gamma}(1)$.($\tilde{\gamma}$ denotes the lift of $\gamma$).
b)If $S$ and $S'$ are isometric, the isometry being $f$. Fix points $x_0\in S$, $y_0\in p^{-1}(x_0)$, $x_1\in S'$ and $y_1\in p'^{-1}(x_1)$. This induces a map $\tilde{f}$ from $\mathbb{H}$ to itself as follows: Consider any point $y\in \mathbb{H}$ and take a curve $\gamma\in \pi_1(S,x_0)$ such that the lift starting at $y_0$ ends at $y$. Define $\tilde{f}(y):= \tilde{\alpha}(1))$ where $\alpha = f\circ \gamma$. This clearly a local isometry as you said. I will prove that it is bijective also.
First i will prove that it is injective. Suppose $\tilde{f}(z)=\tilde{f}(z')=y$. Let $p(z)=x$ and $\tilde{\gamma}$ be any curve from $z$ to $z'$ and $\gamma:= \tilde{\gamma}$. Now, $f(\gamma)\in \pi_1(S',f(x))$ and $\tilde{f(\gamma)}$ starts and ends at $y$ hence $f{(\gamma)}$ is the identity in $\pi_1(S',f(x))$. But a homeomorphism between two spaces induces an isomorphism between their fundamental groups. Hence $\gamma$ has to be identity, which implies that $x=x'$.
Surjectivity:Consider a $y\in \mathbb{H}$. If $x=p'(y)$ then there exists atleast one $y'\in p^{-1}(x)$ such that $y'=\tilde{f}(z)$ for some $z$, as otherwise $x$ will not be in the image of $f$. Fix this $y'$. Take an element $\gamma'\in\pi_1(S',x)$ such that $\tilde{\gamma'}(1)=y$. If $\gamma$ is defined as $f^{-1}(\gamma')$ then $\tilde{f}(\tilde{\gamma}(1))=y$.
Pardon me if i have made some silly error.
