There are likely prettier ways of going about it, but this was what I saw:
Starting with distribution of disjunction over conjunction ($\vee$ over $\wedge$) on the first expression yields $[(P_2 \wedge \neg P_0) \vee P_0] \wedge [(P_2 \wedge \neg P_0) \vee P_1],$ and distributing the other way yields $(P_2 \vee P_0) \wedge (P_2 \vee P_1) \wedge (\neg P_0 \vee P_1).$
Rewriting this as $(P_0 \to P_1) \wedge (\neg P_0 \to P_2) \wedge (\neg P_2 \to P_1),$ it should be clear that the second expression is a logical consequence of the first by simplification. (aka $\wedge$-elimination) Now we just need to show that the implication goes the other way, which is fairly simple.
$1. \textbf{Assume}\ (P_0 \to P_1) \wedge (\neg P_0 \to P_2)$
$\ \ \ 2. P_0 \to P_1$ ($1,$ simplification)
$\ \ \ 3. \neg P_0 \to P_2$ ($1,$ simplification)
$\ \ \ 4. \neg P_2 \to P_0$ ($3,$ contrapositive)
$\ \ \ 5. \neg P_2 \to P_1$ ($2,4,$ hypothetical syllogism)
$\ \ \ 6. (P_0 \to P_1) \wedge (\neg P_0 \to P_2) \wedge (\neg P_2 \to P_1)$ ($1, 5, \wedge$-introduction)
$7. [(P_0 \to P_1) \wedge (\neg P_0 \to P_2)] \to [(P_0 \to P_1) \wedge (\neg P_0 \to P_2) \wedge (\neg P_2 \to P_1)]$ ($1,6,\to$-introduction)
So, because the implications in both directions are tautologies, the material equivalence is a tautology and the statements are logically equivalent.
Best Answer
If you formalise these two statements in first order logic, they read something like:
$\forall f \cdot F(f) \to S(f)$
$\exists f \cdot F(f) \land \lnot S(f)$
where $F(f)$ asserts that $f$ is a fruit and $S(f)$ asserts that $f$ is sweet. The logical negation of 1 is $\lnot\forall f \cdot F(f) \to S(f)$ which (in classical logic) is logically equivalent to 2. Likewise the logical negation of 2 is equivalent to 1.