morphisms representable by algebraic spaces vs morphisms representable by schemes

To answer question 2, the best example I know is $\mathscr{M}_1$, the stack of (proper smooth geom. connected) curves of genus 1. Indeed, Raynaud has contructed an elliptic curve $E\to S$ over a scheme $S$ and an $E$-torsor $X\to S$ which is (an algebraic space but) not a scheme.

This implies two things. First, in order to define $\mathscr{M}_1$ we are forced to take "curve" to mean "algebraic space in curves": if we insist that curves must be schemes, the resulting $\mathscr{M}_1$ will not be a stack for the flat topology, because the above $X$ is locally a (projective) scheme for the flat (even étale) topology on $S$.

Concerning the diagonal, put $I:={\underline{\mathrm {Isom}}}_{\mathscr{M}_1}(E,X)$. There is a monomorphism $X\to I$ (in fact, a closed immersion) identifying $X$ with the subsheaf of $E$-torsor isomorphisms. So, $I$ is not a scheme because $X$ isn't. Viewed as a morphism $S\to \mathscr{M}_1$, $I$ is the pullback of the diagonal under the morphism $S\to \mathscr{M}_1\times\mathscr{M}_1$ given by $(E,X)$. Hence the diagonal is not representable in the scheme sense.

This may not directly answer your question. However, one reason to introduce the notion of an algebraic space is that sometimes it is easier to show that the diagonal is representable by an algebraic space than by a scheme. Consider the following example. Let $G$ be a smooth algebraic group over a field $k$. Consider the classifying stack BG that parametrizes principal bundles that are locally trivial in the fpqc topology. Since $G$ is smooth over a point, and smooth surjective morphisms \'{e}tale locally admit a section, we note that these bundles are a fortiori trivial in the \'{e}tale topology.

Let $T$ be a scheme and suppose we have a morphism $$T \stackrel{(P_1, P_2)}{\longrightarrow} BG \times BG.$$ Here the $P_i$'s are two principal bundles over the scheme $T$. The fiber product $$T \times_{BG \times BG} BG \cong \underline{\operatorname{Isom}}_T(P_1, P_2)$$ where the right hand side is the functor of isomorphisms between $P_1$ and $P_2$. Now choose an \'{e}tale cover $T' \to T$ that trivializes both the $P_i$'s (for instance if $T_i \to T$ trivializes $P_i$ then we can just take $T' := T_1 \times_{T} T_2$). Hence $$\underline{\operatorname{Isom}}_T(P_1, P_2) \times_T T' \cong \underline{\operatorname{Isom}}_{T'}(G_{T'}, G_{T'}) \cong h_{G \times T'}.$$ This shows that the base change of the Isom functor by an \'{e}tale cover is a scheme, from which we conclude that Isom is an algebraic space.