I will focus my answer on the properties which are true for the finite measure spaces but not $\sigma$-finite ones.
Recall Egoroff's theorem:
Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence of measurable function from $X$ to $\mathbb R$ endowed with the Borel $\sigma$-algebra. If $f_n\to 0$ almost everywhere then for each $\varepsilon>0$ we can find $A_{\varepsilon}\in\mathcal A$ such that $\mu(X\setminus A_{\varepsilon})\leq\varepsilon$ and $\sup_{x\in A_{\varepsilon}}|f_n(x)|\to 0$.
It's not true anymore if $(X,\mathcal A,\mu)$ is not finite. For example, if $X=\mathbb R$, $\mathcal A=\mathcal B(\mathbb R)$ and $\mu=\lambda$ is the Lebesgue measure, taking $f_n(x)=\begin{cases}1&\mbox{ if }n\leq x\leq n+1,\\\
0&\mbox{otherwise},
\end{cases}$ we can see that $f_n\to 0$ almost everywhere, but if $A$ is such that $\lambda(\mathbb R\setminus A)\leq 1$, then $\mu(A)=+\infty$, hence $A\cap [n,n+1]$ has a positive measure for infinitely many $n$, say $n=n_k$, so $\sup_A|f_{n_k}|\geq \sup_{A\cap [n_k,n_k+1]}|f_{n_k}|=1$.
An explanation could be the following: if $(X,\mathcal A,\mu)$ is $\sigma$-finite, $\{A_n\}$ is a partition of $X$ into finite measure sets, and a sequence converges almost everywhere on $X$, then we have the convergence in measure on each $A_n$: for $k$ and for a fixed $\varepsilon>0$ we can find a $N(\varepsilon,k)\in\mathbb N$ such that $\mu(\{|f_n|\geq \varepsilon\}\cap A_k)\leq \varepsilon$ if $n\geq N(\varepsilon,k)$. The problem, as the counter-example show, is that this $N$ cannot be chosen independently of $k$.
An other result:
Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence which converges almost everywhere to $0$. Then $f_n\to 0$ in measure.
We can use the same counter-example as above.
Inclusions between $L^p$ space may change whether the measured space is finite. If $(X,\mathcal A,\mu)$ is a finite measured space, then for $1\leq p\leq q\leq \infty$ we have $L^q(X,\mathcal A,\mu)\subset L^p(X,\mathcal A,\mu)$, as Hölder's inequality shows. But with $X=\mathbb N$, $\mathcal A=2^{\mathbb N}$ and $\mu$ the counting measure, we have for $1\leq p\leq q\leq \infty$, $\ell^p\subset l^q$, so the inclusions are reversed.
I am answering my own question because I discovered the answer in my notes, and it may help someone else (it will definitely help me as I return to reference this page in the future).
Here is what is meant by $\sigma$-finiteness being "hidden" in the hypotheses of Fubini's theorem.
To prove Fubini's theorem, we assumed $f \in L^{1}(d\lambda)$. Notice that $f = f\chi_{ \{ x \mid f(x) \neq 0 \} }$, where $\chi_{A} = \begin{cases} 1 & x \in A \\ 0 & x \not \in A \end{cases}$.
Then that means $\int \limits_{X} f \,d\lambda = \int \limits_{X} f\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$. And $\int \limits_{X} f\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{X} f \,d(\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda)$ (for a proof, see the answer and comments here). The measure given by $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$ is $\sigma$-finite, because:
$X = \bigcup \limits_{n = 1}^{\infty} \{ x \mid |f| \geq \frac{1}{n} \} \bigcup \{ x \mid f = 0 \}$.
Now we just need to show each of these sets in the countable union has finite measure with respect to the measure $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$.
For each $n$, we have $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda( \{ x \mid |f| \geq \frac{1}{n} \}) = \int \limits_{ \{ x \mid |f| \geq \frac{1}{n} \} } 1 \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \}} \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \} \cap \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \} } \,d\lambda \leq \int \limits_{ X \times Y } n |f| \,d\lambda < \infty $
since $f \in L^{1}(d\lambda)$. So for each $n$, $ \{ x \mid |f| \geq \frac{1}{n} \} $ has finite measure with respect to the measure $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$. Also, it's clear that $\{ x \mid f(x) = 0 \}$ has measure $0$ with respect to this measure.
So, $(X, \Sigma, \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda)$ is $\sigma$-finite.
Best Answer
I think the trick to this question is to realize that the intuitive answer of defining $\lambda(E)=\mu(E)-\nu(E)$ would not be well defined as, one might end up with $\infty-\infty$.
Given, $\nu$ is $\sigma-$finite, $\exists F_{i},\ \nu(F_{i})<\infty,\ ,\cup_{i}F_{i}=X$. Now, we can also formulate a partition $\{E_{i}\}_{i\in\mathbb{N}}$ of $X$ by $E_{1}=\emptyset,\ E_{i}=F_{i}\backslash\cup_{j<i}F_{j}$, $\mu(E_{i})\leq\mu(F_{i})<\infty$. We will approach the proof by contradiction. Let there exist measures $\lambda_{1},\lambda_{2}$ which both satisfy $\mu=\nu+\lambda_{i}$, but, they disagree on a set $E,$ i.e, $\lambda_{1}(E)\ne\lambda_{2}(E)$.
Now, $\lambda_{i}(E)=\lambda_{i}(\cup_{i}(E\cap E_{i}))=\sum_{i}\lambda_{i}(E\cap E_{i})$. So, the $\lambda_{i}$'s must disagree on atleast one of the $E\cap E_{j}$. Now, by $\sigma$-finiteness of $\nu$, $\nu(E\cap E_{j})<\infty$. Hence, $\lambda_{1}(E\cap E_{j})\ne\lambda(E\cap E_{j})$, but, $\nu(E\cap E_{j})+\lambda_{1}(E\cap E_{j})=\nu(E\cap E_{j})+\lambda_{1}(E\cap E_{j})$, which is a contradiction.
Now, to see why $\sigma-$finiteness of $\nu$ was crucial for our result, let us define \begin{eqnarray*} \nu(E)=\mu(E) & = & 0\ \ \text{if }E=\emptyset\\ & = & \infty\ \mathrm{otherwise}. \end{eqnarray*}
Now, for arbitrary $x\in X$, and $\forall m>0,$ $\mu(E)=\nu(E)+\lambda(E)$ $ $ and \begin{eqnarray*} \lambda(E) & = & 0\ \ \text{if }x\notin E\\ & = & \infty\ \mathrm{otherwise}. \end{eqnarray*} If, X is singleton, then, we would need : \begin{eqnarray*} \lambda(E) & = & 0\ \ \text{if }x\notin E\\ & = & M\ \mathrm{otherwise\ where\ M\in \mathbb{R}}. \end{eqnarray*}
I am still stuck at part (2).