1.Let us consider the set $F[x]_{n,k}$ of all polynomials over $F$ of degree $n$ having preciesly $k$ irreducible factors. The question is: Fix some $n$. Let $p$ be a big prime number.How the cardinality of the sets $\mathbb F_p[x]_{n,k}, 1\leq k\leq n$ looks like? Is it true that $|\mathbb F_p[x]_{n,1}|>0.99\cdot \sum\limits_{k=1}^n |\mathbb F_p[x]_{n,k}|$ when $p$ is very big?
Can we compute the limit of $\dfrac{|\mathbb F_p[x]_{n,j}|}{ \sum\limits_{k=1}^n |\mathbb F_p[x]_{n,k}|}$ for fixed $j,n$ when $p\to +\infty$?
No.
In fact, there is a formula (in terms of mobius inversion and the mobius function $\mu$) for the number of monic irreducibles of degree $n$ in $\mathbb{F}_p$:
$$ \frac{1}{n} \sum_{d \mid n} \mu(d) p^{n/d} $$
You can see here for a more detailed discussion of this formula.
In the case $n$ is prime, this has a simpler closed form (since the divisors $d \mid n$ are easy to understand):
$$ \frac{p^n - p}{n} $$
(again, see here for a more detailed discussion).
Since there are $p^n$ many monic polynomials of degree $n$, this tells us that the fraction of irreducible polynomials (if we continue assuming $n$ is prime, for simplicity) is:
$$\frac{1}{n} \left ( 1 - \frac{1}{p^{n-1}} \right )$$
Which, for large $p$, is roughly $1/n$.
With a little bit more work, you can use this same formula in order to compute the asymptotic size of your sets of interest.
I hope this helps ^_^