Here is a proof for 2. First of all, you can take a vector field $V_0$ in $M$ with isolated zeros as shown here. The idea of the proof is basically "pushing all the zeros to infinity".
Take a covering of the manifold $M$ by compacts $C_n$, $n\geq1$, with $C_n$ contained in the interior of $C_{n+1}$ $\forall n$. We´ll also need $M\setminus C_n$ not to have any component which is relatively compact in $M$. You can achieve this by adjoining to $C_n$ all the relatively compact components of $M\setminus C_n$ (the resulting set is compact, see this).
Now we are going to inductively define a field $V_n$ that has no zeros in $C_n$. To do that, we take the vector field $V_n=(\phi_n)_*V_{n-1}$, where:
By induction hypothesis, $V_{n-1}$ has isolated zeros, and none of them are in $C_{n-1}$.
$\phi_n:M\to M$ is an diffeomorphism fixing $C_{n-1}$ and taking all the (finitely many) zeros of $V_{n-1}$ that were inside $C_n$ outside of it. (To make sure this exists you need the condition ($*$), which implies that every point outside $C_{n-1}$ is connected by a path in $M\setminus C_{n-1}$ to some point outside $C_n$, so you can take it outside $C_n$ by a diffeo fixing $C_{n-1}$).
Now as the sequence $V_n$ of vector fields is eventually constant in any compact subset of $M$, you can consider its limit $V$, which is equal to $V_n$ in $C_n$ so it can´t have any zeros in $C_n$ $\forall n$, meaning it has no zeros.
For 3.1 (or 3.2) to be true it is of course necessary that $\partial M$ has a tangent (transverse) vector field without zeros. This condition will also be sufficient if there is a vector field on $M$ which is tangent (transverse) to $\partial M$ and with isolated zeros outside of $\partial M$, because then we can apply the same proof as for $2)$. There surely is some version of the transversality theorem which implies the existence of such a vector field (if someone provides a reference that would be great). In that case, 3.1 only depends on the topology of $\partial M$ and 3.2 only depends on whether there is a non trivial section of the normal bundle of $\partial M$.
Addendum: A more detailed explanation of how to construct the homeos $\phi_n$ answering Math Dealer's comment:
Note that there is a diffeo $f:\mathbb{R}^d\to\mathbb{R}^d$ taking $0$ to some other point and fixing everything except a compact subset. Thus, by symmetry, $\forall p,q\in\mathbb{R}^d$ there is a diffeo taking $p$ to $q$ and fixing everything except a compact set. So we have
Lemma: For every $d$-manifold $M$, every open $U\subseteq M$ homeo to $\mathbb{R}^n$ and every $p,q$ in $U$ there is a diffeo $f:M\to M$ taking $p$ to $q$ and fixing everything except some compact contained in $U$.
Now in our case, let $M$ be our manifold of dimension $\geq2$, let $P_1,\dots,P_k$ be the zeros of $V_n$ inside $C_n$ (so, inside $C_n\setminus C_{n-1}$). Let $Q_1,\dots,Q_k$ be points such that for all $i$ $Q_i$ is in $ M\setminus C_n$ and in the same connected component of $M\setminus C_{n-1}$ as $P_i$.
Then we can construct (e.g. by recursion) a sequence of disjoint smooth arcs $\gamma_i:[0,1]\to M\setminus C_{n-1}$ with nonzero derivative such that $\gamma_i(0)=P_i$ and $\gamma_i(1)=Q_i$. Now take disjoint nhoods $U_i$ of the arcs $\gamma_i$ (also assume $U_i$ are disjoint with $C_{n-1}$), and then by the lemma there is a diffeo $f:M\to M$ taking $P_i$ to $Q_i$ and fixing everything except $U_1,\dots,U_k$.
