Comparing the rates of hydrolysis of different haloalkanes
This confusion arise because OP did not look at the energy profile of $\mathrm{S_N}$1 mechanism. The first order kinetics of $\mathrm{S_N}$1 reactions (e.g., reactions given in the question) suggests a two-step mechanism in which the rate-determining step consists of the ionization of the substrate (here, it is an alkyl halide), as shown in the following diagram:

The first step, $\ce{R3C-X &lt;=&gt; R3C+ + X-}$, which is the ionization of alkyl halide, is the slowest and the rate determining step. It is true that this step is reversible and is in equilibrium.
However, the next step(s), $\ce{R3C+ + H2O &lt;=&gt; R3C-O+H2 -&gt; R3C-OH + H+}$, is fast and gives the intended product(s). As shown in the energy profile of this mechanism, a carbocation is formed as a high-energy intermediate in first step, and this intermediate bonds immediately to nearby nucleophiles (either $\ce{X-}$ or $\ce{H2O}$ in this case). The rate of this latter reactions would be depends on the values of $\mathrm{\Delta E_\ce{X}}$ (reversible reaction of first step) and $\mathrm{\Delta E_\ce{2}}$ (the second step, $\ce{R3C+ + H2O &lt;=&gt; R3C-O+H2}$), respectively. The value of $\mathrm{\Delta E_\ce{X}}$ may depend on how fast $\ce{X-}$ could take away from the electrophile, meaning the rate of solvation of the respective halide ions come to play here. However, since this is an example of hydrolysis reaction, protic solvent molecules are much more around the electrophile than halide ions, and this competitive reverse reaction is minimal. It is also worth noting that final step, $\ce{R3C-O+H2 -&gt;[X-] R3C-OH + HX}$ is irreversible (not shown in the profile) and would have gone towards to give products eventually.
Therefore, the rate is only depend on the first step. Such reactions would be considered unimolecular and would follow a first-order kinetics. That is the reason to call it $\mathrm{S_N}$1 reactions and having undergo with relevant mechanism. The first step depends on the leaving group ability of the halide. As you learn in the subject, the leaving group ability increase in halogens when going down in the periodic table: $\ce{I- \gt Br- \gt Cl- \gt F-}$ similar to $\ce{C-X}$ bond dissociation energy. Thus, rate of hydrolysis of alkyl halides are:
$$\ce{R3C-I \gt R3C-Br \gt R3C-Cl \gt R3C-F}$$
Hence, $k_1 \gt k_2 \gt k_3$.

Note: Your teacher is correct of the hydration rates of halide ions: $\ce{I- \lt Br- \lt Cl- \lt F-}$ (Ref.1)

References:

Kenzo Hiraoka, Susumu Mizuse, and Shinichi Yamabe, &quot;Solvation of halide ions with $\ce{H2O}$ and $\ce{CH3CN}$ in the gas phase,&quot; J. Phys. Chem. 1988, 92(13), 3943–3952 (https://doi.org/10.1021/j100324a051).