They are all equivalent. The answer to your other question is: the Hamiltonian approach usually works best.
Geodesics can be defined a few ways, since the connection of spacetime is taken to be Levi-Civita.
Let $(\mathcal{M},g)$ denote spacetime $\mathcal{M}$ with metric $g$ and $\gamma:\mathbb{R}\to M,t\mapsto \gamma(t)$ be a curve on $\mathcal{M}$. If $\nabla$ is the Levi-Civita connection on spacetime, then the geodesic equation is
$$\nabla_{\dot\gamma}\dot\gamma=0,$$
where $\dot\gamma$ is the tangent vector of $\gamma$. Some manipulations can bring this into the form given in the OP. On the other hand, we can define the length functional
$$\ell [\gamma]:=\int_a^b\sqrt{-g(\dot\gamma,\dot\gamma)}\,\mathrm{d}t.$$
(The negative appears because $\gamma$ is usually taken to be timelike for GR purposes.) Then it may be shown that
$$\frac{\delta\ell}{\delta\gamma}=0\Longleftrightarrow\nabla_{\dot\gamma}\dot\gamma=0.$$
So the geodesic problem of GR is actually an Euler-Lagrange type problem with Lagrangian $L=\sqrt{-g(\dot\gamma,\dot\gamma)}$.
We can also get a Hamiltonian approach. We first note that the energy functional
$$E[\gamma]=\frac{1}{2}\int_a^bg(\dot\gamma,\dot\gamma)\,\mathrm{d}t$$
has identical Euler-Lagrange equations as the length functional. Proof: Let $D$ be the operator $$Df=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial f}{\partial \dot x}-\frac{\partial f}{\partial x}$$
where $f=f(x,\dot x,t)$, so that $Df=0$ is the Euler-Lagrange equation for $f$. Then a short computation shows that $D\ell=\ell\cdot DE$, so $D\ell=0\implies DE=0$. We note that $H:=\frac{1}{2}g(\dot\gamma,\dot\gamma)$ is reminiscent of the $\frac{1}{2}m\mathbf{v}^2$ term of the classical free particle Hamiltonian. (Hence the name energy functional.)
We can do this more fancily on the cotangent bundle $T^*\mathcal{M}$. Locally trivialize the cotangent bundle in a chart, so we have coordinates $(x^\mu,p_\mu)$. Then put
$$H(x,p):=\frac{1}{2}g^{\mu\nu}(x)p_\mu p_\nu.$$
The Hamiltonian equations
$$\dot x^\mu=\frac{\partial H}{\partial p_\mu},\quad \dot p^\mu=-\frac{\partial H}{\partial x^\mu}$$
are equivalent to the geodesic equation.
The flow obtained from these equations is a Hamiltonian flow.
It turns out that the first Hamilton approach is very useful. It is quite easy to vary the integral $E[\gamma]$. This, combined with the Killing method for obtaining first integrals of the geodesic equation, make for a better strategy than computing the Christoffel symbols and solving the equations outright.
It should be noted that there is another method for solving the geodesic equations: apply the methods of Hamilton-Jacobi theory to the Hamiltonian system described above. This method is used in N. Straumann, General Relativity (2013) to find the geodesics of Kerr spacetime. See also V.I. Arnold, Mathematical Methods of Classial Mechanics (1989) for more on Hamiltonian systems in general.
It should be noted that the Lagrangian and Hamiltonian described above are not actually Legendre transforms of each other.
As you may know, the geodesic equation, your equation (1), is not obtained as the Euler-Lagrange equations of the curve-length functional (2), but rather as the Euler-Lagrange equations of the energy functional
$$E = \frac12\int d\lambda\, g_{\mu\nu} \dot{x}^\mu\dot{x}^\nu.$$
I'm writing $\lambda$ rather than $\tau$ to avoid the suggestion that this has to be the proper time.
It is not very hard to show that extremals of $E$ are extremals of $L$, but the converse doesn't hold, in fact, length extremizing curves are extrema of $E$ if and only if they are true geodesics, i.e. affinely parameterized.
So, your equation (1) are the Euler-Langrange equations of $E$, whose solutions already are affinely parameterized. Adding (3) to it, the only additional requirement is for the curve to be timelike.
All three classes of geodesics, timelike, spacelike and lightlike, have affine parameterizations. For timelike geodesics proper time can be taken as an affine parameter, for spacelike geodesics proper length can be taken, and for lightlike curves no affine parameter has a special meaning.
Best Answer
The Lagrangian does not depend explicitly on the "time" parameter you use to define $\dot{x}$. So, the Hamiltonian function is conserved along the solutions of the E.L. equation. In this case, the Hamiltonian function coincides with the Lagrangian one which, in turn, is the Lorenzian squared norm of the tangent vector. In other words, the type of geodesic is decided on giving the initial tangent vector: its nature (spacelike, timelike, lightlike) is automatically preserved along the curve.