1. Find the smallest prime $p$ that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
It has been verified that $2,3,5,7,11,13,17,19,23,29,31$, and 37 can all be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers:
$$
\begin{array}{l}
2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, \\
7=2^{3}-3^{0}, 11=3^{3}-2^{4}, 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, \\
29=2^{5}-3^{1}, 31=2^{5}-3^{0}, 37=2^{6}-3^{3} .
\end{array}
$$

Pig 41 is the smallest prime number that cannot be expressed in this form. To verify this conjecture, we examine the following two indeterminate equations:
( I ) $2^{u}-3^{v}=41 ;$
( II ) $3^{x}-2^{y}=41$.
Let $(u, v)$ be a non-negative integer solution to equation (I), then $2^{u}>41, u \geqslant 6$. Therefore,
$$
-3^{v} \equiv 1 \text { (mod } 8 \text { ). }
$$

However, the residue of $3^{v}$ modulo 8 can only be 1 or 3, so equation (I) has no non-negative integer solutions.
Let $(x, y)$ be a non-negative integer solution to equation (II), then $3^{x}>41, x \geqslant 4$. Therefore,
$$
2^{y} \equiv 1(\bmod 3) \text {. }
$$

Thus, $y$ must be even. Let $y=2t$, and we get
$$
3^{x} \equiv 1(\bmod 4) .
$$

This implies that $x$ must also be even. Let $x=2s$. Therefore,
$$
41=3^{x}-2^{y}=3^{2s}-2^{2t}=\left(3^{s}+2^{t}\right)\left(3^{s}-2^{t}\right) .
$$

For this equation to hold, it must be that
$$
\left\{\begin{array}{l}
3^{s}+2^{t}=41, \\
3^{s}-2^{t}=1 .
\end{array}\right.
$$

This implies $3^{s}=21, 2^{t}=20$.
But this is impossible. Therefore, equation (II) also has no non-negative integer solutions.
In conclusion, we find that the smallest prime number that cannot be expressed in the form $\left|3^{a}-2^{b}\right|$ (where $a$ and $b$ are non-negative integers) is 41.