Ad (1): It is perfectly true that the tensor product of two reduced algebras over a perfect field is reduced. You can find a proof in Bourbaki's Algebra (Chapter V; §15; 4,5), but of course this venerable author is not especially known for his enthusiasm toward constructive mathematics .

Ad (2): I don't know.

Ad (3): The relation you are looking for is the simplest possible, namely $$Spec(A\otimes_k B)=Spec(A)\times_k Spec(B).$$ But that doesn't help because the product of connected schemes has no reason to be connected. I can only give you the following sufficient condition for connectedness of tensor products of algebras.

Suppose $k \subset K$ is a separable extension of fields such that $k$ is algebraically closed in $K$. Such an extension is called (rather unimaginatively) a REGULAR extension.With this definition, we can state the

Theorem: If $k \subset K$ is regular, then for every $k$-subalgebra $A$ of $K$ and every $k$-algebra $B$ (not related at all to $K$), the tensor product $A\otimes_k B$ is a domain and in particular has connected spectrum.

Here are examples of regular extensions :

a)Every purely transcendantal extension of $k$ is regular.

b)If $k$ is algebraically closed, every extension of $k$ is regular.

PS: Separable extension above means universally reduced and, of course, does not imply that the extension is algebraic.

Ad (1): It is perfectly true that the tensor product of two reduced algebras over a perfect field is reduced. You can find a proof in Bourbaki's Algebra (Chapter V; §15; 4,5), but of course this venerable author is not especially known for his enthusiasm toward constructive mathematics .

Ad (2): I don't know.

Ad (3): The relation you are looking for is the simplest possible, namely $$Spec(A\otimes_k B)=Spec(A)\times_k Spec(B).$$ But that doesn't help because the product of connected schemes has no reason to be connected. I can only give you the following sufficient condition for connectedness of tensor products of algebras.

Suppose $k \subset K$ is a separable extension of fields such that $k$ is algebraically closed in $K$. Such an extension is called (rather unimaginatively) a REGULAR extension.With this definition, we can state the

Theorem: If $k \subset K$ is regular, then for every $k$-subalgebra $A$ of $K$ and every $k$-domain $B$ (not related at all to $K$), the tensor product $A\otimes_k B$ is a domain and in particular has connected spectrum.

Here are examples of regular extensions :

a)Every purely transcendantal extension of $k$ is regular.

b)If $k$ is algebraically closed, every extension of $k$ is regular.

PS: Separable extension above means universally reduced and, of course, does not imply that the extension is algebraic.