= Solution
\b[Pointwise positive Ricci curvature does not imply <compactness> or a finite diameter.] The <complete positively curved paraboloid> $z=x^2+y^2$ has induced metric
$$
g=(1+4r^2)dr^2+r^2d\theta^2
$$
and <Gaussian curvature>
$$
K(r)=\frac{4}{(1+4r^2)^2}>0.
$$
For example, this follows from the graph formula $K=\det D^2z/(1+|Dz|^2)^2$. The metric dominates the Euclidean plane metric, so it is complete by <upward stability of Riemannian completeness>. Its diameter is infinite since its distance dominates Euclidean distance. On a surface the <Ricci tensor> is $Kg$, so it is strictly positive everywhere, but has no uniform positive lower bound.
\b[There is no positive-Ricci metric on $S^2\times S^1$.] If one existed, <compactness> of the <unit tangent bundle> would give $\operatorname{Ric}_{\mathrm{tr}}\geq cg$ for some $c>0$. The lifted metric on the <universal cover> is complete by Q2 and satisfies the same positive lower bound. Part (a) makes this cover <compact>. A covering fiber is closed and discrete and therefore finite, implying a finite <fundamental group>. But
$$
\pi_1(S^2\times S^1)\cong\pi_1(S^1)\cong\mathbb Z,
$$
which is infinite. This is the <finite fundamental group from a uniform positive Ricci bound> obstruction; it rules out every possible metric, not just the standard product metric.
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