Hints:
For (1): To show that a map is continuous you show that the preimage of a closed set is closed. The closed sets in $\mathrm{Spec} \ R$ are all of the form $V(I) = \{\mathfrak p \in \mathrm{Spec} \ R \ | \ I \subseteq \mathfrak p\}$ where $I \subseteq R$ is an ideal. So you show that $\alpha^\ast$ is continuous let $I \subseteq R$ be an ideal and let $\alpha(I)S$ be the ideal in $S$ generated by the image of $I$. You must show that $(\alpha^\ast)^{-1}(V(I)) = V(\alpha(I)S)$.
There's quite a bit of unpacking of definitions to be done here, but it's a very healthy exercise. If you want to work with Spec it really helps to be fluid at unpacking those definitions and that's something that will only come with practice.
For (2): You have to show that it's bijective and you have to show that it's a closed map. For both of these you will heavily use the correspondence theorem for ideals in a localization. Note that an ideal intersects $\{1, f, f^2, \ldots\}$ if and only if it contains $f^n$ for some $n$ and a prime ideal contains $f^n$ if and only if it contains $f$.
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=\operatorname{Spec}A$ and $Y=\operatorname{Spec}B$; one can prove that the function
$$
\Phi:(f,f^{\sharp})\in\operatorname{Hom}_{\bf LocRingSp}((X,\mathcal{O}_X),(Y,\mathcal{O}_Y))\to f^{\sharp}(Y)\in\operatorname{Hom}_{\bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{\sharp})$ is uniquely determined by $f^{\sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{\sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{\sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{\sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=\operatorname{Spec}A$; one can prove that the function
$$
\Phi:(f,f^{\sharp})\in\operatorname{Hom}_{\bf Sch}(Y,X)\to f^{\sharp}(X)\in\operatorname{Hom}_{\bf Ring}(A,\mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $\mathcal{A}$ be a quasi-coherent $\mathcal{O}_S$-algebra; one can prove that the function
$$
\Phi:(f,f^{\sharp})\in\operatorname{Hom}_{S-\bf Sch}(X,\operatorname{Spec}\mathcal{A})\to f^{\sharp}(\operatorname{Spec}\mathcal{A})\in\operatorname{Hom}_{\mathcal{O}_S-\bf Alg}(\mathcal{A},p_{*}\mathcal{O}_X)
$$
is bijective; where $\operatorname{Spec}\mathcal{A}$ is the relative spectrum of $\mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $\operatorname{Spec}(A\otimes_{\mathbb{C}}B)$ is canonically isomorphic to $\operatorname{Spec}A\times_{\operatorname{Spec}\mathbb{C}}\operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
Best Answer
$\text{ker}(\phi)=\bigcap_{P\in Im(\psi)}P$ is not true in general but you can still prove that $V\left(\text{ker}(\phi)\right)=V(\bigcap_{P\in Im(\psi)}P)$.
Note that $Im(\psi)=\{ \phi^{-1}(Q) : Q\in \operatorname{Spec(S)} \}$ so $$V\left(\bigcap_{P\in Im(\psi)}P\right) = V\left(\bigcap_{Q\in \operatorname{Spec(S)}}\phi^{-1}(Q) \right) = V\left( \phi^{-1} \left(\bigcap_{Q\in \operatorname{Spec(S)}} Q \right) \right) =V\left( \phi^{-1}(\operatorname{Nil(S)}) \right)$$ where $\operatorname{Nil(S)}$ is the nilradical of $S$.
Now we can show that $V\left( \phi^{-1}(\operatorname{Nil(S)}) \right)=V\left(\text{ker}(\phi)\right)$. One inclusion is immediate: $V\left( \phi^{-1}(\operatorname{Nil(S)}) \right)\subseteq V\left(\text{ker}(\phi)\right)$ since $\text{ker}(\phi) \subseteq \phi^{-1}(\operatorname{Nil(S)})$.
For the other inclusion, let $P\in V\left(\text{ker}(\phi)\right)$ and we want to show that $\phi^{-1}(\operatorname{Nil(S)}) \subseteq P$. In fact, take an element $x\in \phi^{-1}(\operatorname{Nil(S)})$ then $\phi(x^n)=0$, which implies that $x^n\in \text{ker}(\phi) \in P$, hence, $x\in P$.