$1$. Define $ \begin{align} E^>&=\{x\in X:|f(x)|\ge1\}\\ E^<&=\{x\in X:0<|f(x)|<1\}\\ E^=&=\{x\in X:|f(x)|=0\} \end{align}\tag{1a} $
On $E^>$, $|f(x)|^p$ decreases to $1$ as $p$ decreases to $0$; on $E^<$, $|f(x)|^p$ increases to $1$ as $p$ decreases to $0$; and on $E^=$, $|f(x)|=0$ as $p$ decreases to $0$.
Therefore, by monotone convergence on $E^<$ and dominated convergence on $E^>$, $ \begin{align} \lim_{p\to0^+}\int_{E^>}|f(x)|^p\,\mathrm{d}x&=\mu(E^>)\\ \lim_{p\to0^+}\int_{E^<}|f(x)|^p\,\mathrm{d}x&=\mu(E^<)\\ \lim_{p\to0^+}\int_{E^=}|f(x)|^p\,\mathrm{d}x&=0 \end{align}\tag{1b} $ Summing these yields $ \lim_{p\to0^+}\int_X|f(x)|^p\,\mathrm{d}x=\mu(\{x\in X:|f(x)|\not=0\})\tag{1c} $
$2$. Preliminaries
For $p\gt0$ and $t\ge0$, define $ g_p(t)=\frac{t^p-1}{p}\tag{2a} $ Claim: $g_p(t)$ is non-decreasing in both $p$ and $t$.
$g_p(t)$ is non-decreasing in $t$: This follows from $ g_p^\prime(t)=t^{p-1}\ge0\tag{2b} $
$g_p(t)$ is non-decreasing in $p$: As Didier commented, this follows from $ g_p(t)=\int_1^tu^{p-1}\,\mathrm{d}u\tag{2c} $ and because $u^{p-1}$ is non-decreasing in $p$ when $u\ge1$ and non-increasing in $p$ when $0\le u\le1$.
Furthermore, L'Hopital says $ \lim_{p\to0^+}g_p(t)=\log(t)\tag{2d} $
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Jensen's Inequality says that $h(p)=\|f\|_p$ is non-decreasing in $p$.
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Consider an $\epsilon$ neighborhood of $-\infty$ to be $(-\infty,-\frac1\epsilon)$ and let $L=\lim\limits_{p\to0^+}\log(h(p))$.
For any $\epsilon>0$, choose $q>0$ so that $\log(h(q))$ is within an $\frac{\epsilon}{2}$ neighborhood of $L$.
Choose $r>0$ so that $g_r(h(q))$ is within an $\epsilon$ neighborhood of $L$.
If $p<\min(q,r)$, then both $\log(h(p))$ and $g_p(h(p))$ will be within an $\epsilon$ neighborhood of $L$. Therefore, $ \lim_{p\to0^+}\log(h(p))=\lim_{p\to0^+}g_p(h(p))\tag{2e} $
Main Result
Define $E=\{x:|f(x)|>1\}$, then the results above yield $ \begin{align} \lim_{p\to0^+}\log\left(\|f\|_p\right) &=\lim_{p\to0^+}\frac{\|f\|_p^p-1}{p}\\ &=\lim_{p\to0^+}\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}x\\ &=\color{#C00000}{\lim_{p\to0^+}\int_{E}\frac{|f(x)|^p-1}{p}\,\mathrm{d}x} +\color{#00A000}{\lim_{p\to0^+}\int_{X\setminus E}\frac{|f(x)|^p-1}{p}\,\mathrm{d}x}\\ &=\color{#C00000}{\int_{E}\log|f(x)|\,\mathrm{d}x} +\color{#00A000}{\int_{X\setminus E}\log|f(x)|\,\mathrm{d}x}\\ &=\int_{X}\log|f(x)|\,\mathrm{d}x\tag{2f} \end{align} $ The left limit, in red, is by Dominated Convergence, while the right limit, in green, is by Monotone Convergence. Exponentiate to get $ \lim_{p\to0^+}\|f\|_p=e^{\int_{X}\log|f(x)|\,\mathrm{d}x}\tag{2g} $