Yes, such embeddings of von Dyck groups (and more generally, of discrete isometry groups of hyperbolic space) can be constructed with the standard machinery from algebraic groups and lattices. Here is a sketch of the construction, but for the sake of simplicity, I'll describe the orthogonal case and only describe the case of the quadratic field. The whole construction, with attributions, is explained well in Lubotzky's book. You can treat it as a black box that allows you to embed a subgroup of a group of orthogonal matrices with signature $(n,1)$ and entries in a quadratic field $K$ into the real orthogonal group $O(n+1).$

Let $q$ be the diagonal quadratic form $\sum_{i=1}^{n+1} a_i x_i^2,$ where $a_i$ belong to a real quadratic field $K$ and all $a_i$s are totally positive and $G=O(q)$ is the isometry group of $q$ viewed as an algebraic group. Denote by $\sigma$ the non-trivial Galois automorphism of the field $K.$ The group $G(K)$ of $K$-points of $G$ embeds into $O(n)\times O(n,1).$ In a plain language, $g\in G(K)$ is a matrix with entries in $K$ that preserves the form $q,$ and it is mapped into the pair of real matrices  $(g,g^{\sigma}),$ where $g$ preserves a positively definite quadratic form $q$ and $g^{\sigma}$ preserves the quadratic form $q^{\sigma}$ of signature $(n,1).$ More precisely, by the restriction of scalars from $K$ to $\mathbb{Q}$, $G$ embeds into a product of orthogonal groups over $\mathbb{Q}$ with real forms $O(n+1)$ and $O(n,1),$ and this embedding leads to an embedding $f$ of $G(K)$ into the product of real groups $O(n+1)\times O(n,1).$ Moreover, the composition of $f$ with the projection onto the first factor is an embedding $f'$ of $G(K)$ into $O(n+1).$

Although von Dyck groups are not necessarily arithmetic, they are defined over $\bar{\mathbb{Q}},$ and some of them over quadratic fields. Thus you can embed them into $SO(3),$ although, of course, the image is dense.

A standard example is $q=\sum_{i=1}^n x_i^2 +\sqrt{5}x_{n+1}^2,$ with $K=\mathbb{Q}(\sqrt{5}).$ The conjugate form $q^{\sigma}= \sum_{i=1}^n x_i^2 -\sqrt{5}x_{n+1}^2,$ and it's clear that $q$ is positively definite and $q^\sigma$ has signature $(n,1).$ I think, although I haven't checked the details, that the case $n=2$ takes cares of the von Dyck group $(2,5,5).$

Yes, such embeddings of von Dyck groups (and more generally, of discrete isometry groups of hyperbolic space) can be constructed with the standard machinery from algebraic groups and lattices. Here is a sketch of the construction, but for the sake of simplicity, I'll describe the orthogonal case and only describe the case of the quadratic field. The whole construction, with attributions, is explained well in Lubotzky's book (see also Borel-Wallach). You can treat it as a black box that allows you to embed a subgroup of a group of orthogonal matrices with signature $(n,1)$ and entries in a quadratic field $K$ into the real orthogonal group $O(n+1).$

Let $q$ be the diagonal quadratic form $\sum_{i=1}^{n+1} a_i x_i^2,$ where $a_i$ are non-zero elements of a real quadratic field $K$, and $G=O(q)$ be the isometry group of $q$ viewed as an algebraic group. Denote by $\sigma$ the non-trivial Galois automorphism of the field $K$ and assume that $a_i$ is totally positive for $1\leq i\leq n,$ but $a_{n+1}$ is not. The group $G(K)$ of $K$-points of $G$ embeds into $O(n)\times O(n,1).$ In a plain language, $g\in G(K)$ is a matrix with entries in $K$ that preserves the form $q,$ and it is mapped into the pair of real matrices  $(g,g^{\sigma}),$ where $g$ preserves a positively definite quadratic form $q$ and $g^{\sigma}$ preserves the quadratic form $q^{\sigma}$ of signature $(n,1).$ More precisely, by the restriction of scalars from $K$ to $\mathbb{Q}$, $G$ embeds into a product of orthogonal groups over $\mathbb{Q}$ with real forms $O(n+1)$ and $O(n,1),$ and this embedding leads to an embedding $f$ of $G(K)$ into the product of real groups $O(n+1)\times O(n,1).$ Moreover, the composition of $f$ with the projection onto the first factor is an embedding $f'$ of $G(K)$ into $O(n+1).$

Although von Dyck groups are not necessarily arithmetic, they are defined over $\bar{\mathbb{Q}},$ and some of them over quadratic fields. Thus you can embed them into $SO(3),$ although, of course, the image is dense.

A standard example is $q=\sum_{i=1}^n x_i^2 +\sqrt{5}x_{n+1}^2,$ with $K=\mathbb{Q}(\sqrt{5}).$ The conjugate form $q^{\sigma}= \sum_{i=1}^n x_i^2 -\sqrt{5}x_{n+1}^2,$ and it's clear that $q$ is positively definite and $q^\sigma$ has signature $(n,1).$ I think, although I haven't checked the details, that the case $n=2$ takes cares of the von Dyck group $(2,5,5).$