HALF CLIQUE - NP Complete Problem
Judging by your description and comments, you might be helped the best by an exact description on how reductions can be used to prove NP-completeness:

A problem is NP-complete iff it is in NP and it is NP-hard. This means that any proof of NP-completeness has two parts: a proof that the problem lies in NP and a proof that it is NP-hard.

For the first part, you have to show that YES-instances can be verified in polynomial time using some suitable certificate. Alternatively, you can show the problem can be solved in polynomial time by a non-deterministic Turing machine, but this is not usually done as mistakes are easily made.

In your case, this comes down to proving that for every graph with a $n/2$-clique, you can find some proof that there is indeed such a clique, such that, armed with such a proof, you can check in polynomial time that there is indeed such a clique.

For the second part, you have to show that the problem is NP-hard. This is in nearly all cases shown by proving that your problem is at least as hard as some other NP-hard problem. If HALF-CLIQUE is as least as hard as CLIQUE, it must also be NP-hard.

You do this by proving a reduction FROM CLIQUE, TO HALF-CLIQUE. You 'reduce' the problem, making it 'easier'. You say "Solving CLIQUE is hard, but I have proven that you only need to solve HALF-CLIQUE to solve CLIQUE". (many people, even experts, occasionally say this the wrong way around :))

There are different types of reductions: the reduction that is most commonly used is the one where you map instances of in this case CLIQUE to instances of HALF-CLIQUE whose size is at most polynomially larger, in polynomial time. This means that if we can solve HALF-CLIQUE, then we can also solve CLIQUE by chaining the algorithm and the reduction.

In other words, we must show that we can solve CLIQUE if we can solve HALF-CLIQUE. We do this by showing that for every instance for CLIQUE, we can design an instance of HALF-CLIQUE such that the instance of CLIQUE is a 'yes' instance iff the instance of HALF-CLIQUE is a 'yes' instance.

The proof therefore starts like this: given a graph $G=(V, E)$, I can create some graph $H=(V&#39;, E&#39;)$ such that $G$ contains a $k$-clique iff $H$ contains a $n/2$-clique. I'll leave this part to you (this is the part that requires creativity, the part that is about the specific problem at hand).