= Solution
Fix the curvature convention
$$
\boxed{R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z}.
$$
To prove <tensoriality>, use $[hX,Y]=h[X,Y]-Y(h)X$ and the connection rules. The two $Y(h)\nabla_XZ$ terms cancel, giving $R(hX,Y)Z=hR(X,Y)Z$. Antisymmetry in $X,Y$ gives linearity over smooth functions in the second input. Expanding the third input gives
$$
R(X,Y)(hZ)=hR(X,Y)Z+\bigl(X(Yh)-Y(Xh)-[X,Y]h\bigr)Z=hR(X,Y)Z.
$$
It is therefore a smooth <tensor> of type $(1,3)$. Lowering the output with $g$ gives the type $(0,4)$ <Riemann curvature tensor> $R_4(X,Y,Z,W)=g(R(X,Y)Z,W)$.
For an independent pair $X,Y$, define <sectional curvature> by
$$
K(\operatorname{span}\{X,Y\})=\frac{g(R(X,Y)Y,X)}{g(X,X)g(Y,Y)-g(X,Y)^2}.
$$
Metric compatibility gives $g(R(X,Y)Z,W)=-g(R(X,Y)W,Z)$ by applying $XY-YX-[X,Y]$ to $g(Z,W)$. Together with antisymmetry in $X,Y$, this shows that replacing the pair by $(aX+bY,cX+dY)$ multiplies both numerator and denominator by $(ad-bc)^2$. Thus the value depends only on the plane. Define <Ricci curvature> by $\operatorname{Ric}(Y,Z)=\sum_i g(R(e_i,Y)Z,e_i)$ for any <orthonormal basis>; a trace is independent of the <orthonormal basis>.
For the <curvature of the round unit sphere>, the outward unit normal is the position vector $p$. The tangential projection of ambient differentiation is <torsion-free> and has <metric compatibility>, so uniqueness identifies it with $\nabla$. Ambient differentiation $D$ satisfies $D_Xp=X$ and
$$
D_XY=\nabla_XY-g(X,Y)p,
$$
since differentiating $g(Y,p)=0$ gives its normal component. The ambient curvature is zero. Take tangential components of $D_XD_YZ-D_YD_XZ-D_{[X,Y]}Z=0$ to obtain
$$
\boxed{R(X,Y)Z=g(Y,Z)X-g(X,Z)Y}.
$$
Hence
$$
\boxed{R_4(X,Y,Z,W)=g(Y,Z)g(X,W)-g(X,Z)g(Y,W)}.
$$
Every two-plane has <sectional curvature> one. Tracing the first formula gives $\operatorname{Ric}(Y,Z)=ng(Y,Z)-g(Y,Z)$, and consequently
$$
\boxed{\operatorname{Ric}=(n-1)g,\qquad\Lambda=n-1}.
$$
Thus the sphere is an <Einstein manifold>. When $n=1$ there are no tangent two-planes, the curvature <tensor> is zero and the same Ricci formula gives zero. The declared slot convention fixes all signs.
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