A free body diagram can be helpful here, to find the direction of all the forces acting on each block.
The acceleration of each block will be equal, since they're connected by the string, so
the 80N force acts to accelerate the total mass (m_{1}+m_{2}). The tension does not effect the acceleration of the total mass, it acts equally at both ends of the rope, so it balances out  it acts to slow the acceleration of the 22kg block, and causes of the acceleration of the 11kg block (again, easy to see with a free body diagram).
Part (a): To find the acceleration, apply Newton's second law,
F = ma where m is the total mass (m_{1}+m_{2}) and F is the 80N force, as that's the net force acting on the total mass.
Once you find the acceleration, you can use it to find the tension, by T = m_{2}a since the only horizontal force acting on m_{2} is the tension (if it's not clear, we're still using F = ma, but in this case F = T). We can also find the tension from m_{1}, but it's easier with m_{2}.
Part (b), we use the same process but have to account for friction, which acts to oppose acceleration. Friction force is given by F_{k} = μ_{k}N = μ_{k}mg (μ_{k} = 0.1) and acts on both blocks.
The net force is now: F  (μ_{k}m_{1}g + μ_{k}m_{2}g) = ma, where F = 80N and m is the total mass.
Once you find the acceleration for part (b), we can find the tension as above, but this time we have both tension and friction acting in opposite directions on m_{2}, so
the net force is: T  μ_{k}m_{2}g = m_{2}a.
11/23/2012

Daniel O.
Comments
Thanks you
I'm having trouble still finding the tension in part a and part b.
Once you find the acceleration, you can substitute that into the equation for tension, along with the values for m2, g, friction, and then solve for T.