For each $n\in\mathbb N_1$, set $E_n=|f|^{-1}[(\frac1n,\infty)]$ and $E_0=\varnothing$. Because $f$ is integrable, $\mu(E_n)<\infty$. Let $F=\bigcup_{n\in\mathbb N_1} E_n$ and $F_n=E_n\setminus E_{n-1}$. These sets form a disjoint partition of $F$ because $E_0\subseteq E_1\subseteq \ldots$, so from the countable additivity we have
$$\infty>\int_F|f|\,d\mu=\sum_{n=1}^\infty\int_{F_n}|f|\,d\mu,$$
which means for any $\varepsilon>0$ there's $n\in\mathbb N_1$ such that
$$\varepsilon>\sum_{k=n+1}^\infty\int_{F_k}|f|\,d\mu=\int_{F\setminus E_n}|f|\,d\mu=\int_{X\setminus E_n}|f|\,d\mu$$
because $f$ is $0$ on $X\setminus F$.
For Question 2
The Lebesgue integrability of nonnegative, measurable $f$ means $\int_Ef < +\infty$.
It is then true by the definition of the integral that there exists a bounded, measurable function $g$ of finite support such that $0 \leqslant g \leqslant f$ and
$$\tag{1}\int_E |f-g| < \frac{\epsilon}{2}$$
For Question 1
Let $E_0 = \text{supp }(g)$. It only remains to produce a simple function $\eta$ with finite support in $E_0$ such that $0 \leqslant \eta \leqslant g$ and
$$\tag{2}\int_{E} |g-\eta| = \int_{E_0} |g-\eta| < \frac{\epsilon}{2}$$
The Simple Approximation Theorem (along with the Dominated Convergence Theorem) gives you everything you need to prove (2). Since $g$ is nonnegative with finite support, there exists an increasing sequence of simple functions $\{\phi_n\}$ with finite support such that $0 \leqslant \phi_n \leqslant g$ and $\phi_n \to g$ pointwise. Reread the statement including special cases and proof of this theorem in Royden.
Since $g$ is integrable, by the Dominated Convergence Theorem we have
$$\lim_{n \to \infty} \int_{E_0} \phi_n = \int_{E_0} g$$
Given $\epsilon > 0$ there exists $N $ such that
$$\int_{E_0} |g - \phi_N| = \int_{E_0} (g - \phi_N)= \int_{E_0}g - \int_{E_0}\phi_N < \frac{\epsilon}{2}$$
Taking $\eta = \phi_N$ proves (2).
For Question 3
It would be better to post this as another question, but here is a sketch.
Since this problem arises in Ch.4 we can assume $E \subset \mathbb{R}$.
It is enough to prove that for a characteristic function $\chi_A$ on a measurable set $A \subset E$, there is a step function $\phi$ such that $\int_E| \chi_A - \phi| < \epsilon$. For any $\delta > 0$, there is an open set $O$ containing $A$ with $m(O \setminus A) < \delta$. As an open set $O$ is a countable union of disjoint open intervals
$$O = \bigcup_{j=1}^\infty(a_j,b_j),$$
we have
$$m(O) = \sum_{j=1}^\infty(b_j-a_j) < m(A) + \delta$$
Construct $\phi$ as the characteristic function of a finite union of a sufficiently large number of the intervals $(a_1,b_1), \ldots ,(a_m,b_m)$. This is a step function with the desired properties.
Best Answer
Just take $g=fI_{|x| \leq N}$ with $N$ sufficiently large. Use DCT to show that $\|f-g\|_p \to 0$ as $N \to \infty$.