Clarification regarding multiple modular exponentiation If the base is same and exponents are different, 
for example: 

$R_1=b^x\bmod{p}$; $R_2=b^y\bmod{p}$; $R_3=b^z\bmod{p}$;
($p$ is large prime (2048 bit); $x$, $y$ and $z$ - 160 bit integers))

To calculate $R_1$, $R_2$ and $R_3$ at the same time, then my question is: 

Does the calculation finish in around one exponentiation or three exponentiations? (I feel it is one exponentiation time)
Here's one approach that, assuming $x, y, z$ are random, takes approximately the same time as two exponentiations.

The general idea is to treat x, y, z as base 4 values, which we scan from the lsbit direction.  We keep running values for $R1, R2, R3$, which are initially 1.  At iteration $i$, we have the value $b^{4^i}$ and $b^{2 \cdot 4^i}$ (which takes two modular squarings to compute); if any of $x, y, z$ have a 3 in the digit $i$ position, we also compute the value $b^{3 \cdot 4^i}$ (this happens with probability 0.578, and takes one modular multiplication when it does).  Then, for each of $x, y, z$, if their digit $d$ is not 0, then we multiply in $b^{d \cdot 4^i}$ into the corresponding $R$ value; for each variable, this takes one modular multiply with probability 0.75, for a total of 2.25 expected multiplies.

If we add everything up, we have 2 modular squarings, and 2.828 modular multiplications per step; for 160 bit exponents, there are 80 steps.  If we remove one squaring (for the initial computation of $b^{4^0}$), and the 3 initial modular multiplications of $R1, R2, R3$, we get a total of 159 squarings, and 223 multiplications; this is in the same ballpark as twice the number of operations a good modular exponentiation for a 160 bit exponent would take.

I suspect that this isn't the best possible; it does show that some speed up can be achieved.  On the other hand, I rather suspect that it'd be impossible to compute all three $R1, R2, R3$ in the same time that it takes to compute one of them; some slow down would appear to be inevitable.