= Solution
For a connected complete $n$-dimensional <Riemannian manifold>, $n\geq2$, the <Bonnet-Myers theorem> states that
$$
\operatorname{Ric}\geq(n-1)kg,\quad k>0\quad\Longrightarrow\quad\boxed{\operatorname{diam}M\leq\frac{\pi}{\sqrt k},\quad M\text{ compact},\quad|\pi_1(M)|<\infty.}
$$
The <fundamental group> conclusion follows by applying the same diameter and compactness result to the complete <universal cover>: its covering fiber is closed and discrete in a compact space, hence finite.
There is an important <Ricci curvature> convention in the printed question. With the standard trace convention used in this wiki, the literal hypothesis $\operatorname{Ric}\geq kg$ yields instead $\operatorname{diam}M\leq\pi\sqrt{(n-1)/k}$. The sphere constants in parts (b) and (c) require <normalized Ricci curvature> $\overline{\operatorname{Ric}}=\operatorname{Ric}/(n-1)\geq kg$, equivalently the stronger trace bound $\operatorname{Ric}\geq(n-1)kg$. We prove those conclusions with this intended normalization below and exhibit counterexamples to the literal trace interpretation. Thus no factor of $n-1$ is silently discarded.
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