Tubular Neighborhood Theorem for $C^1$ Submanifold

The answer to your question depends on whether you are looking for a tubular neighborhood in the general differential topological sense or the more restrictive geometric sense. The answer in the topological sense is Yes, but in the geometric sense the answer in general is No. These two conceptions may coincide for $C^2$ submanifolds, but for $C^1$ submanifolds the construction is a bit more subtle, or non-canonical, as I will elaborate below.

A $C^1$ submanifold $M\subset R^n$ may not have any neighborhoods fibrated by normal vectors. So $C^1$ submanifolds do not in general have canonical tubular neighborhoods generated by the exponential map, and "radius" of the neighborhood does not really make sense; however, they still have a tubular neighborhood in the topological sense, i.e., a $C^1$ embedding of the normal bundle which extends that of $M$, as the zero section of the bundle. These may be constructed as follows.

The fastest way to construct a topological tubular neighborhood for $M$ is by noting that there exists a $C^1$ diffeomorphism $\phi\colon R^n\to R^n$ such that $\phi(M)$ is $C^\infty$ (e.g., see Thm. 3.6 in Chap. 2 of Hirsch's book). Then $\phi(M)$ is going to have a tubular neighborhood $U$ in the standard sense, obtained by applying the inverse function theorem to the exponential map (or taking the union of all sufficiently short normal vectors), and $\phi^{-1}(U)$ yields the desired tubular neighborhood of $M$ (which will be $C^1$-diffeomorphic to the normal bundle).

An example of a $C^1$ submanifold without a neighborhood fibrated by normals may be constructed as follows. Take an ellipse in $R^2$, which is not a circle, and let $M$ be the inner parallel curve of the ellipse which passes through the foci. Then $M$ will be $C^1$, but it will not have a tubular neighborhood in the geometric sense. Indeed, the nearest point projection map into $M$ will not be one-to-one in any neighborhood of $M$.

Yes, you can; you don't need compactness of $N$ nor finite dimension: the key word here is paracompactness. I wish to do a general remark and recall some basic facts, since this seems to be a recurrent topic. Let $N$ be this embedded sumbanifold of $M$. You want a homeomorphism (maybe with more regularity and with some special feature) between the normal bundle $E$ of $N$ in $M$ and some open nbd of your submanifold $N\subset M$, that extends the natural homeomorphism $j$ between the zero-section $E_0$ of $E$, and $N$. The typical situation is that you already have some map $\Phi:E\to M$ that extends $j$ and which is a local homeo at any point of $E_0$ (a local condition that it is usually easy to check e.g. by the local inverse theorem), and you want to show it subordinates a global homeo between an open nbd $U$ of $E_0$ in $E$, and an open nbd $V$ of $N$ in $M$. Clearly, $\Phi$ is an open map on some nbd of $S_0$ so all you need is to have $\Phi_U$ injective on some nbd, which is not completely obvious; then you may further restrict it to an open nbd $U'$ diffeomorphic to $E$. The key lemma to get injectivity is the following topological "extension of homeomorphisms" lemma:

Lemma. Let $E$ and $F$ be paracompact spaces; let $X$ be a closed subset of $E$ and $Y$ a closed subset of $F$. Let $f:E\to F$ be a continuous map such that:

(i) $f$ is a local homeo at any $x\in X$ (meaning that for any $x\in X$ there is an open nbd $B_x$ of $x$ in $E$ such that $f(B_x)$ is open in $F$ and $f_{|B_x}:B_x\to f(B_x)$ is homeo);

(ii) $f_{|X}:X\to Y$ is bijective (hence a homeo).

Then there are open nbds $U$ of $X$ in $E$ and $V$ of $Y$ in $F$ such that $f_{|U}:U\to V$ is a homeo.