= Solution
An <orthonormal coframe> is a collection of independent <differential forms> $E^{\hat a}=e^{\hat a}{}_\mu dx^\mu$ satisfying $g_{\mu\nu}=\eta_{\hat a\hat b}e^{\hat a}{}_\mu e^{\hat b}{}_\nu$, with $\eta=\operatorname{diag}(-1,1,1,1)$. Locally one can diagonalize the <metric tensor> and rescale the resulting covectors, or use <Gram-Schmidt orthogonalization for a symmetric bilinear form>, choosing a timelike direction first. The coframe is not unique: a local Lorentz transformation preserves the metric. For the <Levi-Civita connection>, solve <Cartan's first structure equation> and metric compatibility together:
$$
dE^{\hat a}+\omega^{\hat a}{}_{\hat b}\wedge E^{\hat b}=0,\qquad
\omega_{\hat a\hat b}=\eta_{\hat a\hat c}\omega^{\hat c}{}_{\hat b}=-\omega_{\hat b\hat a}.
$$
These determine the <connection 1 forms>. The antisymmetry applies to the lowered pair of indices; a mixed time-space pair instead has $\omega^0{}_i=\omega^i{}_0$. Next <Cartan's second structure equation> gives the <curvature 2-forms>, and reading off their coefficients determines the <Riemann curvature tensor>:
$$
\Omega^{\hat a}{}_{\hat b}=d\omega^{\hat a}{}_{\hat b}+\omega^{\hat a}{}_{\hat c}\wedge\omega^{\hat c}{}_{\hat b}
=\frac12R^{\hat a}{}_{\hat b\hat c\hat d}E^{\hat c}\wedge E^{\hat d},\qquad
R_{\hat b\hat d}=R^{\hat a}{}_{\hat b\hat a\hat d}.
$$
This also states the curvature-sign convention. Contracting gives the <Ricci tensor>, and its coordinate components are $R_{\mu\nu}=e^{\hat a}{}_\mu e^{\hat b}{}_\nu R_{\hat a\hat b}$.
On a static patch with $V>0$, choose
$$
E^0=\sqrt V\,dt,\quad E^1=\frac{dr}{\sqrt V},\quad E^2=r\,d\theta,\quad E^3=r\sin\theta\,d\phi.
$$
Write $h=\sqrt V$, $a=h'$, $b=h/r$, and $c=\cot\theta/r$. The <exterior derivative> gives
$$
dE^0=-aE^0\wedge E^1,\qquad dE^1=0,\qquad
dE^2=bE^1\wedge E^2,\qquad dE^3=bE^1\wedge E^3+cE^2\wedge E^3.
$$
The nonzero independent <connection 1 forms> are consequently
$$
\omega^0{}_1=aE^0,\qquad \omega^1{}_2=-bE^2,\qquad
\omega^1{}_3=-bE^3,\qquad \omega^2{}_3=-cE^3,
$$
with $\omega^1{}_0=\omega^0{}_1$, spatial mixed pairs antisymmetric, and the remaining forms zero. For example,
$$
\Omega^0{}_1=d(aE^0)=-(hh''+h'^2)E^0\wedge E^1=-\frac{V''}2E^0\wedge E^1.
$$
Similarly $\Omega^0{}_2=\omega^0{}_1\wedge\omega^1{}_2=-abE^0\wedge E^2$, and the angular derivatives in $d\omega^1{}_3$ cancel against $\omega^1{}_2\wedge\omega^2{}_3$. The resulting <Cartan curvature forms for a static spherical metric> are
$$
\boxed{\begin{aligned}
\Omega^0{}_1&=-\frac{V''}2E^0\wedge E^1,\\
\Omega^0{}_2&=-\frac{V'}{2r}E^0\wedge E^2,&\Omega^0{}_3&=-\frac{V'}{2r}E^0\wedge E^3,\\
\Omega^1{}_2&=-\frac{V'}{2r}E^1\wedge E^2,&\Omega^1{}_3&=-\frac{V'}{2r}E^1\wedge E^3,\\
\Omega^2{}_3&=\frac{1-V}{r^2}E^2\wedge E^3.
\end{aligned}}
$$
For the last entry, $d(-cE^3)=E^2\wedge E^3/r^2$ and $\omega^2{}_1\wedge\omega^1{}_3=-VE^2\wedge E^3/r^2$. The remaining <curvature 2-forms> follow from $\Omega_{ab}=-\Omega_{ba}$.
Put $A=V''/2$, $B=V'/(2r)$, and $C=(1-V)/r^2$. Reading the <Riemann curvature tensor> from the forms and contracting gives the diagonal orthonormal <Ricci tensor>
$$
\boxed{R_{\hat a\hat b}=\operatorname{diag}(A+2B,-A-2B,C-2B,C-2B).}
$$
For example $R_{00}=R^1{}_{010}+R^2{}_{020}+R^3{}_{030}=A+2B$, while $R_{22}=-B-B+C$. All off-diagonal entries vanish because the <curvature 2-forms> only involve their own index planes. In the coordinate basis the nonzero components are
$$
\boxed{\begin{aligned}
R_{tt}&=V\left(\frac{V''}2+\frac{V'}r\right),&
R_{rr}&=-\frac1V\left(\frac{V''}2+\frac{V'}r\right),\\
R_{\theta\theta}&=1-V-rV',&
R_{\phi\phi}&=(1-V-rV')\sin^2\theta.
\end{aligned}}
$$
The angular <Einstein field equations> with the negative <cosmological constant> give $(rV)'=1+3r^2/l^2$. Integrating introduces one real constant, conventionally written $-2m$:
$$
\boxed{V(r)=1-\frac{2m}{r}+\frac{r^2}{l^2}.}
$$
Substitution gives $V''/2+V'/r=3/l^2$, so the time and radial <Einstein field equations> hold as well. The solution is the <Schwarzschild--anti-de Sitter black hole> family, including its zero and negative mass limits.
For $m>0$, $V$ increases strictly from $-\infty$ at $r=0$ to $+\infty$ at infinity. Its unique positive simple zero $r_h$ satisfies $r_h^3+l^2r_h=2ml^2$. This zero is a <coordinate singularity>, not a curvature singularity: with the ingoing coordinate $v=t+\int dr/V$, the radial metric is $-Vdv^2+2dv\,dr$ and is regular there. Inside the future black-hole region, $\nabla r=\partial_v+V\partial_r$ is future timelike, so every future-directed causal trajectory decreases $r$ and cannot return to the asymptotic anti-de Sitter boundary. Conversely, every exterior radius admits an outward <radial null geodesic> reaching that boundary, since $\int_r^\infty dr/V$ converges. Thus \b[$m>0$ gives an event horizon at the unique positive zero of $V$]. Here the <event horizon> is defined relative to the asymptotic timelike boundary of <Anti-de Sitter spacetime>. For $m=0$, $V=1+r^2/l^2>0$ and the spacetime is pure <Anti-de Sitter spacetime>, with \b[no black-hole event horizon]. For $m<0$, $V=1+2|m|/r+r^2/l^2>0$, so there is also \b[no event horizon]; the mass term produces a <naked singularity> at $r=0$. The field equation itself does not impose a positive mass integration constant.
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