Easiest Round-off Trick!!!:
Step 1: Look at the exponents of the different alphabets.
Step 2: Is there any relation between the different exponents?
Step 3a: If there is NO relation between the exponents then the language is Regular and Context free. eg: L= $a^n b^m$ , m,n >=0
Step 3b: If there are relations between the exponents, for instance, $L=a^m b^n$, and the relation can be m>n or m!=n or m< n.
You will need one counter here to keep a count, inrementing it by 1 for n times for a and compare with b by decrementing by 1 for n-times and you get a 0 for a match(assuming, m=n). These languages are Not Regular but context free and accepted by a PDA.
Step 4: If you have more than one relation or need more than one counter, for instance, $L= a^m b^n c^k$, m=n=k. Here, you need 2 such counters. First, count all a's and copy this count value to another counter. Then, compare with b by decrementing it n-times to get a 0. Again, decrementing the other counter n-times to compare it with c and accept the language if both the counters are 0.
These languages are neither Regular nor Context Free but they are Context Sensitive and thus Recursive, although, all Recursive languages are not Context Sensitive