Is Cohen immersion conjecture (theorem) known for vector bundles? R. Cohen proved the immersion conjecture in a 1985 Annals paper:
Cohen, Ralph L., The immersion conjecture for differentiable manifolds, Ann. Math. (2) 122, 237-328 (1985). ZBL0592.57022.
Any smooth compact n-dimensional manifold admits an immersion into Euclidean space of dimension 2n-a(n), where a(n) is the number of 1's in the binary decomposition of n.
Is there any result of this kind for (the total space of) a vector bundle E over compact manifold? Notice that the sphere bundle of E is compact. Maybe there is a silly argument...
 A: The answer is no. 
Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle.  This is a considerably better number than $2n-a(n)$. 
You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds. 
Consider for example the line bundles over surfaces. The immersion conjecture would say they all immerse in $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$ but I think we can argue they immerse in $\mathbb R^4$.  
Said another way, we are asking what line bundles are sub-bundles of normal bundles of immersed surfaces in $\mathbb R^4$.   For orientable surfaces you can quickly generate any isomorphism type.  For non-orientable surfaces I believe you can, as well.  You immerse the surface in $\mathbb R^3$ and use the $1$-cocycle as a way to guide how you "flip" the bundle, when thought of as inside the larger $\mathbb R^4$.
A: I don't know if the following remarks can be considered "of this kind".
Let $E$ be the total space of a rank $r$ vector bundle over the compact $n$-manifold $M$. Then, as you said, by Cohen $M\subseteq \mathbb{R}^{2n-a(n)}$. 
Also (see Atiyah, K-theory, Corollary 1.4.14) you can realize $E$ as a sub-vector-bundle of a trivial bundle $E\subseteq \underline{\mathbb{R}}^m$ for a suitable $m$. In fact, looking at the proof of the preceding Lemma 1.4.12 it seems* to me that you can take $m$ to be $\leq r\times t(E)$ where $t(E)$ is the minimum number of elements of an open cover of $M$ which is trivializing for $E$. And from this, it seems that $t(E)\leq n+1$. 
So, you could embed $E$ in $\mathbb{R}^{2n-a(n)+r\cdot (n+1)}$. 
$^*$ You have maps of vector bundles $\theta_\alpha:{\underline{\mathbb{R}}}_{U_\alpha}^r\to E|_{U_\alpha}$ which you glue by a partition of unity $\{p_\alpha\}$, to obtain that $E$ is generated by global sections, by the map $\theta=\sum_\alpha p_\alpha \cdot\theta_\alpha:\prod^{t(E)}\underline{\mathbb{R}}^r\to E\to 0$. Since $E$ is isomorphic to its dual, you can dualize the map and obtain an inclusion of $E$ into a trivial v.b.
Remark: I haven't thought if this would give, in the case of a total space of a v.b., a better result than Cohen's embedding theorem itself applied to the projectivization $\mathbb{P}(E\oplus\mathbb{R})$ (or the one-point compactification of $E$ along the fibers) which has dimension $N=n+r$. One should ask if $2n-a(n)+r(n+1)< 2N-a(N)$ for some $n,r$.
