first of all, this belong to beginers/generic programming,please consider posting topics like this there,you will find easier the help you want.
Second answering homework-like questions is not allowed here so i will give you only some pointers.

Considering the bitwise operators in C/C++(i assume you use this language)
it's easy to implement binary masking in a given value and a given bit position.

AND &
OR |
shift right,left >> <<


example:

i want to set bit '3'(right to left) in number n,i know from boolean algebra :

a OR 1 = 1
i also know that shifting a value left by s bits is equal to value * 2^s
(easier to see like a position rather directly a value,ofcourse you can put it like a value directly if you are sure to remember 2^n n=0..32,64 etc.]

so i do : n = n OR 2^(3-1) and it's done

to C language:

n |= (1 << 2);

bit masking is just using AND or OR with a mask

for example if you want to mask 10101010 with 00001111 to set the first 4 bits to 0, use 101010 AND 00001111 and you get 0001010

in what programming language do you have to do this? I'm sure most languages have operators for this.

Bit-wise operations are pretty simple in C. Basically, they are the standard logical operations done on several bits in parallel. So, the AND, OR, XOR, and NOT operators work just like you expect them to (but simultaneously on several bits in parallel).

Now, as far a "bit masking" goes, here are the keys:

• x AND 0 = 0, x AND 1 = x
• x OR 0 = x, x OR 1 = 1
• x XOR 0 = x, x XOR 1 = NOT x

These operations allow you to select bits to be turned off, turned on, or inverted. They also give you a way to select which bits you want to operate on.

For example, suppose you want to compute z = a AND b, but for bit 5 only (and without affecting any of the other bits in z). There are many ways to do it. Here is one way.

    mask = 0x20               // Set bit 5 to 1 and all others to 0 (for later use)             ...0100000     inverted_mask = NOT mask  // Set bit 5 to 0 and all others to 1 (for later use)             ...1011111     z5 = a AND b              // Get the results of a AND b             ...?a?????        AND ...?b?????            ==========            ...?x?????     z5 = z5 AND mask          // Since x AND 1 = x and x AND 0 = 0, bit 5 contains the result, the rest are 0.             ...?x?????        AND ...0100000            ==========            ...0x00000     z = z AND inverted_mask   // Since x AND 1 = x and x AND 0 = 0, bit 5 is 0, the rest are unchanged.             ...???????        AND ...1011111            ==========            ...?0?????     z = z OR z5               // Since x OR 0 = x, bit 5 contains the value of z5, the rest are unchanged             ...?0?????         OR ...0x00000            ==========            ...?x?????