Is the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ right exact? Let $(\mathsf{CRing})^{\text{op}}$ be the category opposite to the category $\mathsf{CRing}$ of commutative rings with one, and $\mathsf{Set}$ the category of sets. Recall that $\text{Spec}(A)$ is the set of prime ideals of $A$, and that a morphism $f:A\to B$ in $\mathsf{CRing}$ induces, in a functorial way, a map $\text{Spec}(B)\to\text{Spec}(A)$ sending $\mathfrak q$ to $f^{-1}(\mathfrak q)$.

(1a) Is the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ right exact?

Here are four other equivalent forms of Question (1a):

(1b) Does the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ commute with finite colimits?

Let $(A_i)_{i\in I}$ be a projective system of commutative rings indexed by a finite category $I$. 

(1c) Is the natural map $$\operatorname*{colim}_i\text{Spec}(A_i)\to\text{Spec}\left(\lim_i A_i\right)$$ bijective?

The fourth and fifth forms will use the fact that our functor commutes with finite coproducts.  (Recall that coproducts in $(\mathsf{CRing})^{\text{op}}$ correspond to products in $\mathsf{CRing}$.)

(1d) Does the functor $\text{Spec}:(\mathsf{CRing})^{\text{op}}\to\mathsf{Set}$ commute with cokernels?

Let $A\rightrightarrows B$ be two parallel morphisms in $\mathsf{CRing}$.

(1e) Is the natural map $$\operatorname{Coker}\Big(\text{Spec}(B)\rightrightarrows\text{Spec}(A)\Big)\to\text{Spec}\Big(\operatorname{Ker}(A\rightrightarrows B)\Big)$$ bijective?

Here is a second question:

(2) In the above setting, is the natural map $$\text{Spec}(A)\to\text{Spec}\Big(\operatorname{Ker}(A\rightrightarrows B)\Big)$$ surjective?

Of course, if the answer to (2) is "not necessarily", then the answer to (1) will be "no".
Martin Brandenburg asked a somewhat related question.
 A: No.  For instance, let $A=B=\mathbb{C}[x]$ and let $f:A\to B$ be the identity map and $g:A\to B$ send $x$ to $x+1$.  Then the equalizer of $f$ and $g$ is the set of polynomials $p$ such that $p(x)=p(x+1)$, which is just the constant polynomials $\mathbb{C}$ (since if such a $p$ were nonconstant, it would need to have infinitely many roots).  But the coequalizer of spectra has more than one point, since it is just the quotient $\mathbb{C}/\mathbb{Z}$ (together with a generic point).
The answer to the second question is also no.  For instance, let $A=\mathbb{C}[x,y,r,s]/(rx+sy-1)$ and $B=\mathbb{C}[x,y,r,s,t,u]/(rx+sy-1,tx+uy-1)$.  There is an obvious inclusion map $f:A\to B$ but there is also a map $g:A\to B$ which sends $r$ and $s$ to $t$ and $u$.  I claim that the equalizer of $f$ and $g$ is just $C=\mathbb{C}[x,y]$.  Given this claim, we see that the prime ideal $P=(x,y)\subset C$ does not extend to any prime ideal of $A$, so the map $\operatorname{Spec} A\to\operatorname{Spec} C$ is not surjective.
To prove the claim, let $a\in A$ be such that $f(a)=g(a)$.  We can write $a$ as a sum of monomials which do not contain $rx$ (since we can replace $rx$ with $1-sy$).  First suppose that no monomial of $a$ contains $r$.  Then $a$ is a polynomial in just $s$, $x$, and $y$; suppose $s$ does appear in $a$.  We can pick values $x_0,y_0\in\mathbb{C}$ with $x_0\neq 0$ such that when we substitute $x_0$ for $x$ and $y_0$ for $y$, $a$ becomes a nonconstant polynomial $p(s)$ in $s$.  For any $\alpha,\beta\in\mathbb{C}$, we can now consider the homomorphism $e_{\alpha,\beta}:B\to\mathbb{C}$ which sends $y$ to $y_0$, $s$ to $\alpha$, $u$ to $\beta$, $x$ to $x_0$, $r$ to $\frac{1-\alpha y_0}{x_0}$, and $t$ to $\frac{1-\beta y_0}{x_0}$.  We then have $e_{\alpha,\beta}(f(a))=p(\alpha)$ and $e_{\alpha,\beta}(g(a))=p(\beta)$.  Since $p$ is nonconstant, we can choose $\alpha$ and $\beta$ such that $p(\alpha)\neq p(\beta)$ and reach a contradiction.
So, we may assume that $r$ does appear in $a$; write $a=\sum_{i=0}^n a_i x^i+\sum_{j=1}^m b_jr^j$ where the $a_i$ and $b_j$ are in $\mathbb{C}[y,s]$.  Replacing every appearance of $r$ by $\frac{1-sy}{x}$, we may consider $a$ as a Laurent polynomial in $x$ with coefficients in $\mathbb{C}[s,y]$ (the division by $x$ is not an issue because we will eventually be substituting a nonzero complex number for $x$).  Since $r$ appears in $a$, some $b_j$ is nonzero.  Thus one of the coefficients of this Laurent polynomial involves $s$, since we replaced $r$ with $\frac{1-sy}{x}$.  So, we can also consider $a$ as a nonconstant polynomial in $s$ with coefficients in $\mathbb{C}[x,1/x,y]$.  We can now choose $x_0,y_0\in\mathbb{C}$ with $x_0\neq 0$ such that when we substitute them for $x$ and $y$, $a$ becomes a nonconstant polynomial in $s$, and reach a contradiction as in the previous case.
