Answer to question 1: Consider de polynomial $(x^2+1)^2\in\Bbb Q[x]$. It has no rational roots and it is reducible over $\Bbb Q$. Your reasoning fails.
The absence of roots only guarantees that there are no factors of degree $1$.
Your polynomial has degree $4$, since it has no rational roots, it has no linear factors (that is factors of degree $1$), but it could have two factors of degree $2$.
Assume it does and try to reach a contradiction. If you do, then it is irreducible over $\Bbb Q$.
Answer to question 2: The factors don't need to be monic polynomials, but they can be. If you find a factorization in which the factors aren't monic, you just have to multiply by a certain constant to make the factors monic.
Answer to question 3: If you find a factorization $(x-\alpha )q(x)$, where $q(x)$ is a polynomial with rational coefficients of degree $3$, then the polynomial will have a rational root, namely $\alpha$ and you have estabalished it doesn't.
Answer to question 4: The coefficients need not be integers, but (the second) Gauss's lemma allows us to assume the coefficients are integers, making the calculations much simpler.
The basic idea is to factor the polynomial over $\mathbb F_p$ for one or more suitable primes $p$, and attempt to lift these factorizations to factorizations over $\mathbb Z$. There are bounds on the size of the primes $p$ one needs to consider, making this an effective algorithm.
Over $\mathbb F_p$ one can first do a number of preliminary reductions, to reduce the problem to one where the polynomial $f$ to be factored is square-free and all irreducible factors are of fixed degree $d$. A simple way to proceed is then the (probabilistic) method of Cantor and Zassenhaus: If one takes a random monic polynomial $g$ of degree $\le 2d-1$, then the gcd of $f$ and $g$ will be a non-trivial factor of $f$ with probability about $1/2$. So, just proceed taking random $g$'s and computing the gcd with $f$, until you have a complete factorization. Other algorithms (e.g., Berlekamp) exist for this final step.
The book A Course in Computational Algebraic Number Theory by H. Cohen treats these problems in Chapter 3.4 (Factorization of Polynomials Modulo $p$) and Chapter 3.5 (Factorization of Polynomials over $\mathbb Z$ or $\mathbb Q$). These two chapters don't require any knowledge of algebraic number theory, and are easy to skim through or read.
Best Answer
Let $f(x) = x^{6} - x^{5} + x^{3} - x^{2} + 1 \in \mathbb{Z}[x]$. Consider the polynomial $g(x) = f(x+2) = 37 + 120x + 165x^2 + 121x^3 + 50x^4 + 11x^5 +x^6$. We note that the coefficients are all positive and bounded by $165$. We have that $g(172) = 27592587407701$ and $g(172)$ is prime. By Cohn's Criterion , $g(x)$ is irreducible in $\mathbb{Z}[x]$, and it follows $f(x)$ is irreducible in $\mathbb{Z}[x]$. A bit overkill, but it gets the job done.