Apologies - a better explanation than I started with - thanks to people for helping. It is obvious that there are many bad cases for rank - the problem is are there a reasonable number of good cases? 


A rank for fgp modules (fgp = finitely generated projective, as appropriate) over a unital C\* algebra $A$ can be given by using a trace (from $A$ to complex numbers) of the matrix trace of the associated projection matrix. The problem is when this is multiplicative for bimodules - i.e. when the rank of $N \otimes\_A M$ is the rank of $N$ times the rank of $M$. (It is likely easier to look at $N \otimes\_A M$ in terms of dual bases rather than the original projection matrices.)


Classically this is just the rank of a vector bundle - and for locally trivial bundles there is no reason why this is multiplicative unless the topological space is connected (just have different local ranks on different components - a trace would somehow average this into a single number). We either need some condition on the bundle over different components, or an assumption that the space is connected.


The problem is to take a suitable unital algebra (with some form of connectedness condition) and show that there is a reasonable class of fgp modules with multiplicative ranks. I guess that $C^\*$ algebras with normalised tracial states and some form of connectedness condition would be good. It is very likely that some regularity condition would be needed on the modules apart from being fgp.


Another way to look at the multiplicative problem: Take a dual basis $e^i\in E$ and $e\_i\in E'$. Then the projection matrix is $P\_{ij}=e\_i(e^j)$. Given a dual basis for $F$, $f^i\in F$ and $f\_i\in F'$, a projection matrix for $F\otimes\_A E$ is $e\_i(f\_j(f^k).e^r)$. Now the map $\mathcal{P}\_E:A\to M\_n(A)$ given by $\mathcal{P}\_E(a)=e\_i(a.e^j)$ is an algebra map, sending 1 to $P$. Given the matrix trace and $\tau$ a trace on $A$, we have a trace $\tau\circ\mathrm{trc}$ on $M\_n(A)$. Now if we had $\tau\circ\mathrm{trc}\circ\mathcal{P}\_E:A\to\mathbb{C}$ being a fixed multiple of $\tau$ I guess we would have multiplicativity... [I seem to be editing my own question again - should I put this in an answer to my own question - I am unused to protocols hereabouts...]


[Another edit.] As far as I can see for multiplicativity, the key is that $\tau\circ\mathrm{trc}(P)$ should be independent of the algebra normalised trace $\tau$ (at least for a large number of traces). Then $\tau\circ\mathrm{trc}\circ\mathcal{P}\_E:A\to\mathbb{C}$ is another (not normalised) trace, and applying this to the projection matrix for the other module should give the result. But what could give $\tau\circ\mathrm{trc}(P)$ independent of $\tau$? Is that the same as saying that just using the relation $\tau(xy)=\tau(yx)$ is enough to get $\tau$ applied to a multiple of the identity?


Remarks 
(1) there is such a rank for objects in a rigid braided monoidal ribbon category (see Majid's book), but that is too much machinery for the present case. 


(2) A useful measure of `nice' modules from Morita theory is that the evaluation map $:E'\otimes\_A E\to A$ is onto - this translates into the existence of suitably sized matrices $U,V$ so that the matrix trace of $UPV$ is the identity in the algebra. 


(3) If there is a projection matrix (equivalently a dual basis) so that the matrix trace of the projection is a multiple of the identity, then (a) the rank is independent of the algebra trace, and (b) I suspect that the multiplicative result is true. However, this condition is likely to be too strong.


(4) By Blackadar's book, for a $C^\*$ algebra $P$ can be chosen to be Hermitian, so for a tracial state the rank is positive. When is the matrix trace of $P$ (a positive algebra element) invertible?



rank of fin gen projective modules over C* algebras
Here is an answer to the second question (item (4) of the list in the question): The set of possible "matrix traces" will contain an invertible element if and only if the projection (or the module) is full. That is, the entries of the projection span the algebra as a two sided ideal in $A$. (This is also equivalent to the condition mentioned in item (2) of the list.)


Let me assume for simplicity that the projection $p\in A$ (rather than $M\_n(A)$). If 
$$
p=\sum\_{i=1}^n x\_i^\*x\_i
$$ and 
$$
a=\sum\_{i=1}^n x\_ix\_i^\*
$$ 
then $a$ is a possible matrix trace for $pA$. Indeed, with 
$X=(x\_1,x\_2,\dots,x\_n)$ we get $p=X^\*X$ and $XX^\*$ has matrix trace equal to $a$.


Now suppose that $p$ spans $A$ as a two-sided ideal. This entails the existence of 
$d\_i\in A$ such that


$$
d\_1^\*pd\_1+d\_2^\*pd\_2+\cdots d\_n^\*pd\_n=1
$$


Let $d=\sum\_{i=1}^n d^\*\_id\_i$, $x\_i=\frac{1}{ \|d\|^{1/2}}d\_ip$, and $y=(1-\frac{d}{\|d\|})^{1/2}p$. Then 
$$
\sum\_{i=1}^n x\_i^\*x\_i+y^\*y=p(\frac{d}{\|d\|})p+p(1-\frac{d}{\|d\|})p=p.
$$ 
On the other hand,
$$
\sum\_{i=1}^n x\_ix\_i^\*+yy^\*=1/\|d\|+(1-d/\|d\|)^{1/2}p(1-d/\|d\|)^{1/2}.
$$
The right hand side is invertible, since we are adding a positive multiple of the identity to a positive element.


This proves the "harder" of the two implications. One can see that the possible matrix traces of a projection are always inside the 2-sided ideal generated by the projection (or its entries in the matrix case), which proves the other implication.


