(c) Calculate the homotopy fibre of the inclusion $i_{X} : X \vee X \rightarrow X \times X. $ Here is the question:
Let $F$ be the homotopy fiber of the inclusion $X \rightarrow X \times X.$ 
(1)Show that $\pi_{i}(F) \cong \pi_{i +1}(X).$
Here is the answer of this part:
Show that $\pi_{i}(F) \cong \pi_{i +1}(X) $ where $F$ is the homotopy fiber of the inclusion $X \rightarrow X \times X.$
(2)Identify the homotopy type of $F.$
Seems like the answer also in the above link but I did not understand that.
(c) Calculate the homotopy fibre of the inclusion $i_{X} : X \vee X \rightarrow X \times X. $ 
I got a hint to use the First Cube Theorem and The Second Cube Theorem . Could anyone show me more details of using them please?
 A: 1+2) By 'the inclusion', I assume you mean the inclusion into one factor, say 
$$in_2:X\hookrightarrow X\times X,\qquad x\mapsto (\ast,x).$$
The easiest way to proceed is to notice that this map is principal. That is, it is the fibre inclusion of the fibration
$$pr_1:X\times X\rightarrow X,\qquad (x,y)\mapsto x.$$
Thus there is a fibration sequences
$$\dots\rightarrow F\rightarrow X\xrightarrow{in_2}X\times X\xrightarrow{pr_1}X$$
where $F$ is the homotopy fibre in question. This makes it clear that
$$F\simeq \Omega X.$$
3) and or c) I assume you mean the wedge $X\vee X$ rather than the smash. Start by realising $X\vee X$ as the pushout 
$\require{AMScd}$
\begin{CD}
\ast @> >> X\\
@VV V   @VVV\\
X @>>> X\vee X.
\end{CD}
Assuming that $X$ is well pointed, this square is also a homotopy pushout. Embed 
$$j:X\vee X\rightarrow X\times X$$
in the standard way, realising the wedge as the subspace of points $\{(x,\ast),(\ast,x)\in X\times X\mid x\in X\}$. 
Now use the embedding $j$ to pull back the path space fibration over the above homotopy pushout to build a homotopy commutative cube. Restricting $j$ along the bottom-right hand legs of the pushout, we get the two inclusions
$$in_1,in_2:X\rightarrow X\times X$$
which we considered above, and we've already identified their homotopy fibres as $\Omega X$. At the back of the top face in the cube we have the homotopy fibre of the constant map $\ast\rightarrow X\times X$, and this is $\Omega(X\times X)\cong \Omega X\times \Omega X$. Thus the top face of the cube looks like
\begin{CD}
     \Omega X\times \Omega X @>pr_2 >> \Omega X\\
    @VV pr_1V   @VVV\\
    \Omega X @>>> F_j.
\end{CD}
where $F_j$ is the homotopy fibre of $j:X\vee X\hookrightarrow X\times X$.
I've labelled the maps in the back of this square because I happen to know what they are. You need to check that these are the correct maps. There is no trick to this, just turn all your maps into fibrations in the standard way and it's fairly clear that these maps are indeed the projections.
Now we can appeal to Math's Cube Theorem, which tells us that the homotopy fibre $F_j$ is the homotopy pushout of
$$\Omega X\xleftarrow{pr_1}\Omega X\times \Omega X\xrightarrow{pr_2}\Omega X.$$
But this is by definition the join $\Omega X\ast\Omega X$, and since we're assuming that $X$ is well-pointed we get the homotopy type as
$$F_j\simeq \Omega X\ast\Omega X\simeq \Sigma\Omega X\wedge \Omega X.$$
