A lot of insight comes from having counterexamples in your head. For example, take the 1st, which is essentially the same as the 4th: I can never remember which way Fatou's lemma goes until I do some simple examples like:
(Counter)example:
Let $f_n(x) = 1/n$ for all $x$. Clearly $\int_{-\infty}^{\infty} f_n(x)dx = \infty$ for all $n$, so its limit is also $\infty$. But $f_n(x)$ converges point wise to $0$, so the integral of the limit is $0$. So we now remember that Fatou's lemma for non-negative functions $f_n(x)$ is:
\begin{eqnarray*}
\liminf_{n\rightarrow\infty} \int f_n \geq \int \liminf_{n\rightarrow\infty} f_n
\end{eqnarray*}
The Lebesgue dominated convergence theorem is the usual thing to apply if you want to get equality results. That is actually proven directly from Fatou: Suppose $f_n(x)$ are non-negative, converge pointwise to $f(x)$, and there is an L1-integrable $y(x)$ such that $f_n(x) \leq y(x)$ for all $x$ and $n$. Then $y(x)-f_n(x)$ is non-negative, so by Fatou:
\begin{eqnarray*}
\liminf_{n\rightarrow\infty} \int (y-f_n) \geq \int (y-f)
\end{eqnarray*}
Since $\int y$ is finite we can subtract it from both sides to get:
\begin{eqnarray*}
\liminf_{n\rightarrow\infty} \int -f_n \geq \int -f
\end{eqnarray*}
Multiplying the above by $-1$ gives:
\begin{eqnarray*}
\limsup_{n\rightarrow\infty} \int f_n \leq \int f
\end{eqnarray*}
On the other hand, by regular Fatou we know for the non-negative function $f_n(x)$:
\begin{eqnarray*}
\liminf_{n\rightarrow\infty} \int f_n \geq \int f
\end{eqnarray*}
The previous two inequalities together imply that the $\limsup$ and $\liminf$ are the same, so:
\begin{eqnarray*}
\lim_{n\rightarrow\infty} \int f_n = \int f
\end{eqnarray*}
Then the general Lebesgue result for functions $f_n(x)$ that are possibly negative uses a bounding function $y(x)$ that satisfies $|f_n(x)|\leq y(x)$ for all $x$, and is proven by breaking $f_n(x)$ into positive and negative parts.
The last one is also directly related to this issue of pushing limits thru integrals. Assume $f(x,t)$ is L1-integrable (over $t$) for all $x$. So we can talk about $\int f(x+h,t)dt - \int f(x,t)dt$ without worrying about pesky cases of $\infty - \infty$. Then, from the definition of derivative we wish to look at:
\begin{eqnarray*}
\lim_{h\rightarrow 0}\int \frac{f(x+h,t)-f(x,t)}{h}dt
\end{eqnarray*}
So we can define $g_h(t) = \frac{f(x+h,t)-f(x,t)}{h}$. If we assume $\lim_{h\rightarrow0}g_h(t) = \frac{\partial}{\partial x}f(x,t)$, then we can use the Lebesgue Dominated Convergence theorem to show $\lim_{h\rightarrow 0} \int g_h(t)dt = \int \frac{\partial}{\partial x}f(x,t)dt$.
This is justified if we can find an L1-integrable bounding function $y(t)$ that satisfies $|g_h(t)| \leq y(t)$ for all $t$ and $h$. Such bounding functions can often be found by exploiting certain properties of $f(x,t)$, such as Lipschitz properties and/or finite support over $t$.
The second property is interesting and had me fooled for a bit! I found a link to a good counterexample here:
Under what conditions can I interchange the order of limits for a function of two variable?
The simplest (not the strongest) sufficient condition is that $f(x,y)$ be continuous at all points in an open neighborhood of $(a,b)$ (including $(a,b)$ itself). You can prove that and it leads to good "geometric" intuition.
A more general sufficient condition for $\lim_{x\rightarrow a} [\lim_{y\rightarrow b} f(x,y)]$ and $\lim_{y\rightarrow b} [\lim_{x\rightarrow a} f(x,y)]$ to exist and be equal is if all of the following three conditions hold:
(i) $f(x,y)$ is continuous at $(a,b)$.
(ii) There is a $c>0$ such that $\lim_{x\rightarrow a} f(x,y)$ exists as a real number whenever $0<|y-b|<c$.
(iii) There is a $d>0$ such that $\lim_{y\rightarrow b} f(x,y)$ exists as a real number whenever $0<|x-a|<d$.
Peter Franek's comment perhaps leads to the most "geometrical" intuition (for the 3rd problem). You can consider functions that have their actual value uniformly pushed down to 0, but they still oscillate wildly.
