2 "k-algebra" replaced by "k-domain"

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 (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 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. 1 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.