Question: Honestly I have no idea where to start and It'd be so nice if someone provided me with a step by step explanation , Thank you !

1) $\neg q \rightarrow \neg p \equiv p \rightarrow q$

2) $p \lor (p \land q) \equiv p$

3) $(p \lor q) \land \neg p = \neg p \land q$

Answer: 
Since $\mathbf{A}\_{k\times n}$ is a k by n matrix ,through your way of picking $X\_{ij}$ :

$$\Pr\left[X\_{ij}=1| j \_ {fixed}\right]=\dfrac{1}{n}$$
And there will be k one's in your Matrix $$\mathbf{X}=[X\_{ij}]$$
So, 
\begin{align}\mathbb{E}\left[X\_j\right]&=\sum\_{j=1}^n\mathbb{E}\left[X\_{ij}\right]\\&=\frac{k}{n},\end{align}
It has nothing to do with how you generate the random matrix A.
Symetry arguments are nice (!). Here, $X\_{ij}=1$ for row-maximums. Assuming an i.i.d. continuous source distribution, there is a single row-maximum per row, for a total of $k$ row-maximums for the whole matrix; i.e.
$$
\sum\_{i,j}X\_{ij}=\sum\_{i}X\_i=k \rightarrow \sum\_{i}E(X\_i)=k.
$$
As symetry tells us that $E(X\_i)$ does not depend on $i$,
$$
k=\sum\_{i}E(X\_i)=n\,E(X\_i)\rightarrow E(X\_i)=\frac kn.
$$
BTW, as each element in a column has a probability $1/n$ of being a row-maximum and that these are independent between rows, the $X\_i$'s each follow a *binomial distribution* with parameters $k$ and $1/n$. This is another way of arriving at $E(X\_i)=k\times \frac 1n$. Of course, the $X\_i$'s are not independent of one another.