What is the difference between goto and longjmp() and setjmp()?

A goto statement implements a local jump of program execution, and the longjmp() and setjmp() functions implement a nonlocal, or far, jump of program execution. Generally, a jump in execution of any kind should be avoided because it is not considered good programming practice to use such statements as goto and longjmp in your program.

A goto statement simply bypasses code in your program and jumps to a predefined position. To use the goto statement, you give it a labeled position to jump to. This predefined position must be within the same function. You cannot implement gotos between functions.

 When your program calls setjmp(), the current state of your program is saved in a structure of type jmp_buf. Later, your program can call the longjmp() function to restore the program’s state as it was when you called setjmp().Unlike the gotostatement , the longjmp() and setjmp() functions do not need to be implemented in the same function. However, there is a major drawback to using these functions: your program, when restored to its previously saved state, will lose its references to any dynamically allocated memory between the longjmp() and the setjmp(). This means you will waste memory for every malloc() or calloc() you have implemented between your longjmp() and setjmp(), and your program will be horribly inefficient. It is highly recommended that you avoid using functions such as longjmp() and setjmp() because they, like the gotostatement, are quite often an indication of poor programming practice.

How does throwing and catching exceptions differ from using setjmp and longjmp?

The throw operation calls the destructors for automatic objects instantiated since entry to the try block.

Exceptions are in the mainstream of C++ now, so most programmers, if they are familiar with setjmp and longjmp, should know the difference. Both idioms return a program from the nested depths of multiple function calls to a defined position higher in the program. The program stack is "unwound" so that the state of the program with respect to function calls and pushed arguments is restored as if the calls had not been made. C++ exception handling adds to that behavior the orderly calls to the destructors of automatic objects that were instantiated as the program proceeded from within the try block toward where the throw expression is evaluated. 
It's okay to discuss the notational differences between the two idioms. Explain the syntax of try blocks, catch exception handlers, and throw expressions. Then specifically address what happens in a throw that does not happen in a longjmp. Your answer should reflect an understanding of the behavior described in the answer just given.
One valid reason for not knowing about exception handling is that your experience is exclusively with older C++ compilers that do not implement exception handling. I would prefer that you have at least heard of exception handling, though.
It is not unusual for C and C++ programmers to be unfamiliar with setjmp/ longjmp. Those constructs are not particularly intuitive. A C programmer who has written recursive descent parsing algorithms will certainly be familiar with setjmp/ longjmp. Others might not, and that's acceptable. In that case, you won't be able to discuss how setjmp/longjmp differs from C++ exception handling, but let the interview turn into a discussion of C++ exception handling in general. That conversation will reveal to the interviewer a lot about your overall understanding of C++.

Besides communication cost, what is the other source of inefficiency in RPC?

context switches, excessive buffer copying.
How can you optimise the communication?
(ans : communicate through shared memory on same machine, bypassing the kernel

Given an eight-bit bitmap graphics file, devise an algorithm to convert the file into a two-bit ASCII approximation

Assume that the file format is one byte for every pixel in the file, and that the approximation will produce one ASCII character of output for each pixel. This problem is easier to solve than it sounds. This is one of the tricks used in technical interview questions. Problems may be obscured or made to sound difficult. Don't be fooled! Take the time to think about the core of the problem. In this case, all you want is an algorithm for reading the values in a file and outputting characters based upon those values.
Eight-bit numbers can be in the range from 0 to 255. Two-bit numbers are in the range from 0 to 3. Basically, we want to divide the 256 numbers specified by an eight-bit number into four ranges, which can be indicated by a two-bit number. So, divide the range of 0 to 255 uniformly into four separate ranges: 0 to 63, 64 to 127, 128 to 191, and 192 to 255.
You then have to assign an ASCII character to each of those four ranges of numbers. For example, you could use "_", "~", "+", and "#". Then, the algorithm is as follows:

1. Open the file.
2. For every byte in the file:
a. Read in one byte.
b. If the value is in the range 0..63, we'll print '_'.
c. If the value is in the range 64..127, we'll print '~'.
d. If the value is in the range 128..191, we'll print '+'.
e. If the value is in the range 192..255, we'll print '#'.
3. Close the file.

Devise an algorithm for detecting whether a given string is a palindrome

For the sake of this problem, assume that the string has been stripped of punctuation (including spaces), and has been converted to a single case. The most efficient way to detect whether a string is a palindrome is to create two pointers. Set one at the beginning of the string, and one at the end. Compare the values at those locations. If the values don't match, the string isn't a palindrome. Otherwise, move each pointer inward and repeat the comparison. Stop when the pointers are pointing to the same position in the string (if its length is an odd-number) or when the pointers have "crossed" (if the string's length is an even-number).

For the sake of this problem, assume that the string has been stripped of punctuation (including spaces), and has been converted to a single case. The most efficient way to detect whether a string is a palindrome is to create two pointers. Set one at the beginning of the string, and one at the end. Compare the values at those locations. If the values don't match, the string isn't a palindrome. Otherwise, move each pointer inward and repeat the comparison. Stop when the pointers are pointing to the same position in the string (if its length is an odd-number) or when the pointers have "crossed" (if the string's length is an even-number).

Given the time, devise an algorithm to calculate the angle between the hour and minute hands of an analog clock

The important realization for this problem is that the hour hand is always moving. In other words, at 1:30, the hour hand is halfway between 1 and 2. Once you remember that, this problem is fairly straightforward. Assuming you don't care whether the function returns the shorter or larger angle.