$解:(1)設(shè)購進(jìn)A品牌農(nóng)產(chǎn)品x箱,購進(jìn)B品牌的農(nóng)$
$產(chǎn)品(100-x)箱.$
$根據(jù)題意,得80x+130(100-x)=10000$
$解得x=60$
$100-60=40(箱)$
$答:購進(jìn)A品牌農(nóng)產(chǎn)品60箱,購進(jìn)B品牌農(nóng)產(chǎn)品$
$40箱.$
$(2)設(shè)購進(jìn)B品牌農(nóng)產(chǎn)品y箱。根據(jù)題意,得$
$(120-80)(100- y)+(200-130)y≥5600$
$解得y≥53 \frac{1}{3}$
$答:至少需購進(jìn)B品牌農(nóng)產(chǎn)品54箱.$