How can a block which is not receiving the direct force have a greater acceleration? 
I solved it like this:
$$F(\text{st max})=5\text{ N}$$
For the top block,
$$\begin{align}
6\text{ N} - 5\text{ N} &= 1a \\
a &= 1\ \mathrm{m/s^2}
\end{align}$$
For the lower block, the driving force will be the frictional force, so
$$\begin{align}
2a &= 5\text{ N} \\
a &= \frac{5}{2} = 2.5\ \mathrm{m/s^2}
\end{align}$$
I am confused as to how the lower block could have acceleration greater than the upper block, since the force is acting on the top block.
 A: Your calculations are wrong. 
The basic assumption that friction = u x N 
u = coefficient of friction
N = Normal force (in this case the weight of the block)
Above assumption is valid only if there is relative motion between the two blocks i.e a case of sliding motion, but before we consider that sliding occurs we should verify whether the block are moving relative to each other of not i.e. checking for static friction.
Now the maximum value static friction can reach is uN i.e. Sliding/kinetic friction but can also be lesser than that. 
Taking that into account and assuming friction to be f (a variable) and no relative motion between the blocks. 
No relative motion means that both blocks will have same acceleration.
Calculations :
6N− f =a m/s2 (for small block)
f = 2a m/s2 (for big block)
substituting f=2a for small block
6N - 2a = a m/s2
6N = 3a m/s2
2 m/s2 = a
both blocks having same acceleration, hence no relative motion.
Value of friction in this condition is 2 x 2 = 4N which is less than uxN = 5N
A: You have got the right answers, just are interpreting it a little bit wrong;   
The net force on upper block is $6N - 5N = 1N$ and it's accelerations is $1 ms^{-2}$ while the force on bottom block is $5N$ and it's acceleration is $2.5ms^{-2}$  but you are missing the fact that the acceleration of the bottom block is with respect to ground while that of upper block is with respect to lower block.  Therefore, if you see from ground bottom block accelerates with $2.5ms^{-2} $ while upper block moves with $3.5ms^{-2} $.
I apologise for misunderstanding and posting the wrong answer earlier
A: This the result of not keeping track of what your calculation or measurement is relative to. The small block is not accelerating very much relative to the large block, this is correct:
5N-6N=-1(kg)*1(m/s^2)
The large block is accelerating much relative to the surface, this is correct:
2(kg)*(5/2)(m/s^2)=5N
But how is the small block moving relative to the surface?
It is accelerating much faster than the large block:
(5/2)(m/s^2)+1(m/s^2)=(7/2)(m/s^2) or as you say (7/2)a
This is the acceleration of the small block you are missing out on.
Yes relative to the large block the small block is only accelerating at "a" or 1(m/s^2).
But to find this relative to the surface you must add the acceleration of the large block.
