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Let $G$ be a simple connected graph.

1) If $G$ has set of degrees of vertices that contains only 3 numbers then there exists a path of length $3$ in which every vertex has different degree.

2)If $G$ has set of degrees of vertices that contains only 4 numbers does it contains a path of length $4$ in which every vertex has different degree?

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1 is wrong. Let $G = (\{\,1, 2, 3, 4\,\}, \{\,\{\,1, 2\,\}, \{\,2, 3\,\}\,\})$. Then $\{\,\deg v\colon v \in V(G)\,\} = \{\,0, 1, 2\,\}$. Also there is a connected graph on $9$ vertices with set of degrees $\{\,1, 2, 3\,\}$ that doesn't satisfy the first condition.

2 is wrong for the same reason.

Even more, I state that for any $k$ and $\ell$ there is a graph with cardinality of set of vertex degrees equal to $k$ that doesn't contain path of length $\ell$ with vertices of all possible degrees.

Edit. Ok, I have smaller connected graph. Let $G = (\{\,1, 2, 3, 4, 5, 6, 7\,\}, \{\,\{\,1, 2\,\}, \{\,2, 3\,\}, \{\,3, 4\,\}, \{\,4, 5\,\}, \{\,5, 6\,\}, \{\,5, 7\,\}, \{\,6, 7\,\}\,\})$. The set of vertex degrees is $\{\,1, 2, 3\,\}$, graph is connected.

Let $G = (\{\,1, 2, 3, 4, 5, 6, 7, 8\,\}, \{\,\{\,1, 2\,\}, \{\,2, 3\,\}, \{\,3, 4\,\}, \{\,4, 5\,\}, \{\,5, 6\,\}, \{\,5, 7\,\}, \{\,5, 8\,\}, \{\,6, 7\,\}, \{\,6, 8\,\}\,\})$. The set of vertex degrees is $\{\,1, 2, 3, 4\,\}$.

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    Thnx i would prefer examples for connected graphs as it is stated. – 2017-02-09
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    Yes i got it you are right – 2017-02-09
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    i think its enough 6 and 7 vertices – 2017-02-09
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    It depends on measure of path length. I used length measured in number of edges. If you measure length as number of vertices, then you can remove one vertex per graph. – 2017-02-09