Here is a partial result that says that if the sphere can be partitioned into small congruent polygons, then the polygons must be "thin" in some sense. Specifically:
Theorem Suppose a sphere can be partitioned into congruent spherical $n$-gons of diameter $\varepsilon>0$ in an edge-to-edge tiling. Then the area $A$ of the $n$-gon is $O_n( \varepsilon^{2 + c_n} )$ for some constant $c_n>0$ depending on $n$.
In particular, the area is much less than the disk of radius $\varepsilon$, when $\varepsilon$ is small; geometrically, this asserts that the $n$-gon is "thin" compared to its circumscribing disk. This is broadly consistent with the examples provided by the OP of tiling by thin triangles.
Probably the restriction on edge-to-edge tiling can be removed by a variant of the argument. In the convex case, Euler characteristic arguments give $n \leq 5$ as noted already in comments, while the triangular tiling case has already been classified, so it "only" remains to rule out the case of small thin quadrilaterals and small thin pentagons, which seems doable in principle (the thinness does place significant constraints on the angles and lengths of these $n$-gons, which would be difficult to reconcile with the tiling hypothesis, especially if one imposes an edge-to-edge tiling condition. For instance, a thin pentagon cannot have all five sidelengths equal to each other.) but there appears to be quite a lot of combinatorial cases to check.
Proof We can take $\varepsilon$ to be small, and write the area $A$ as $\delta \varepsilon^2$. We allow all implied constants to depend on $n$. Let $r$ be a real number with $1 \ll r \ll 1/\varepsilon$, and consider which copies of the $n$-gon intersect a disk of radius $r \varepsilon$. Let $N_r$ denote the number of copies that intersect the disk, and $M_r$ the number that intersect but are not contained in the disk. Standard volume packing arguments (similar to those used to establish the trivial bound of $\pi r^2 + O(r)$ in the Gauss circle problem of estimating the number of lattice points in a disk of radius $r$) show that $N_r \asymp r^2/\delta$ and $M_r = O(r/\delta)$. (Because we are in the regime $r \ll 1/\varepsilon$, the fact that we have a spherical geometry instead of a plane geometry does not cause significant distortion to these estimates.) On the other hand, from Gauss-Bonnet, the angles $\alpha_1,\dots,\alpha_n$ of each $n$-gon add up to $(n-2)\pi + \delta \varepsilon^2$. Summing up over all the $N_r$ copies of the $n$-gon and double-counting, we conclude that
$$ N_r ((n-2)\pi + \delta \varepsilon^2) = 2\pi V_r + \sum_{i=1}^n c_{r,i} \alpha_i$$
where $V_r$ are the number of vertices of the packing that lie inside the disk, and $c_{r,i}$ are the number of angles $\alpha_i$ corresponding to one of the $M_r$ boundary $n$-gons that lie outside the disk. Clearly the $c_{r,i}$ are natural numbers with $0 \leq c_{r,i} \leq M_r$. In particular
$$ 2V_r - (n-2)N_r = O( N_r \delta \varepsilon^2 + M_r )$$
$$ = O( r^2 \varepsilon^2 + r/\delta ) = O(r/\delta)$$
since $r \ll 1/\varepsilon \leq \frac{1}{\delta \varepsilon^2}$. We have thus created linear relations
$$ N_r = \sum_{i=0}^n c_{r,i} \beta_i \tag{1}$$
for any $1 \ll r \ll 1/\varepsilon$, where $\beta_i = \alpha_i / \delta \varepsilon^2$ for $i=1,\dots,n$, $\beta_0 = \pi / \delta \varepsilon^2$, and $c_{r,0} = 2V_r - (n-2)N_r$. Thus the $c_{r,i}$ are integers of size $O(r/\delta)$ and $N_r$ is an integer of magnitude $\asymp r^2/\delta$. The key point here is that the integer $N_r$ is significantly larger than any of the coefficient integers $c_{r,i}$. This turns out to be exploitable information even without knowing much about the magnitudes of the variables $\beta_i$.
Now let $1 < r_1 < r_2 < \dots < r_{n+2}$ be a sequence of real numbers defined recursively by $r_1 := C$ and
$$ r_{j+1} = C r_j^2 r_{j-1} \dots r_1 / \delta^j$$
for $j=1,\dots,n+1$, where $C$ is a large constant to be chosen later. Then
$$ r_{n+2} \asymp_C \delta^{-O(1)}.$$
If $r_{n+2} \geq 1/\varepsilon$ then $\delta = O( \varepsilon^{c_n} )$ for some $c_n>0$ and we are done, so we may suppose that $r_{n+2} < 1/\varepsilon$. In particular we have the linear relations
$$ N_{r_j} = \sum_{i=0}^n c_{r_j,i} \beta_i \tag{2}$$
for $j=1,\dots,n+2$.
It turns out that this system of $n+2$ linear equations on the $n+2$ variables $1, \beta_0,\dots,\beta_{n+1}$ is non-singular and so (2) cannot be satisfied (it would imply $1=0$, which is absurd). To show this, we first claim inductively that for each $1 \leq j \leq n+2$, there is a $j \times j$ minor
$$ \begin{pmatrix}
N_{r_1} & c_{r_1,i_1} & \dots & c_{r_1,i_{j-1}} \\
\vdots & \vdots & \ddots & \vdots \\
N_{r_j} & c_{r_j,i_1} & \dots & c_{r_j,i_{j-1}}
\end{pmatrix}
$$
which is nonsingular for some $0 \leq i_1 < \dots < i_{j-1} \leq n$. This is clear for $j=1$ since $N_{r_1}$ is non-zero. Now suppose the claim is already established for some $1 \leq j \leq n$. Using (2) and multilinearity of the determinant in the columns, we conclude that there is a $j \times j$ minor
$$ \begin{pmatrix}
c_{r_1,i'_1} & \dots & c_{r_1,i'_j} \\
\vdots & \ddots & \vdots \\
c_{r_j,i'_1} & \dots & c_{r_j,i'_j}
\end{pmatrix}
$$
which is nonsingular for some $0 \leq i'_1 < \dots < i'_{j+1} \leq n$. The determinant is obviously an integer. By cofactor expansion, the $j+1 \times j+1$ minor
$$ \begin{pmatrix}
N_{r_1} & c_{r_1,i'_1} & \dots & c_{r_1,i'_j} \\
\vdots & \vdots & \ddots & \vdots \\
N_{r_{j+1}} & c_{r_{j+1},i'_1} & \dots & c_{r_{j+1},i'_j}
\end{pmatrix}
$$
then has determinant of magnitude
$$ \gg N_{r_{j+1}} - O( \sum_{i=1}^j N_i \prod_{1 \leq k \leq j+1: k \neq i} (r_k/\delta) )$$
$$ \gg r_{j+1}^2/\delta - O( r_{j+1} r^2_j r_{j-1} \dots r_1 / \delta^{j+1} )$$
$$ > 0$$
by the construction of $r_{j+1}$ (if $C$ is large enough), closing the induction.
Applying the inductive claim at $j=n+2$, we see that the $n+2 \times n+2$ matrix
$$ \begin{pmatrix}
N_{r_1} & c_{r_1,0} & \dots & c_{r_1,n+1} \\
\vdots & \vdots & \ddots & \vdots \\
N_{r_{n+2}} & c_{r_{n+2},0} & \dots & c_{r_{n+2},n+1}
\end{pmatrix}
$$
is nonsingular, but this contradicts (2) as discussed previously. $\Box$
UPDATE: The following recent papers, when put together, form a complete classification of monohedral edge-to-edge tilings of the sphere by convex polygons:
Wang, Erxiao; Yan, Min, Tilings of the sphere by congruent pentagons. I: Edge combinations $a^2b^2c$ and $a^3 bc$, Adv. Math. 394, Article ID 107866, 36 p. (2022). ZBL1484.52016.
Wang, Erxiao; Yan, Min, Tilings of the sphere by congruent pentagons. II: Edge combination $a^3b^2$, Adv. Math. 394, Article ID 107867, 68 p. (2022). ZBL1484.52017.
Yohji Akama, Erxiao Wang, Min Yan, Tilings of the Sphere by Congruent Pentagons III: Edge Combination $a^5$
Hoi Ping Luk, Min Yan, Tilings of the Sphere by Congruent Pentagons IV: Edge Combination $a^4b$
Ho Man Cheung, Hoi Ping Luk, Min Yan, Tilings of the Sphere by Congruent Quadrilaterals or Triangles
Most of the tilings in the classification have a bounded number of faces (and hence diameter bounded below), but there are some families of tilings with unboundedly many faces, mostly based around the "earth map" tiling construction, which when unwrapped looks something like one of these two figures (taken from Figure 3 of https://arxiv.org/abs/2307.11453):
In each of these tilings, the upper end converges to the north pole, the bottom end converges to the south pole, and the left and right ends are glued together. The "height" of the tiling is bounded, but the "width" is unbounded. In these constructions, the individual n-gons have small area, but still large diameter (it only takes a bounded number of them to reach from one pole to another); they are quite thin as the number of faces goes to infinity, consistent with the above result. So there are no counterexamples to Q1 of the OP arising from edge-to-edge tilings of convex polygons. I suspect that some of the examples in the linked papers may also provide a positive answer to Q2 or Q3, though one would have to check each of the tilings in these papers separately to do so. (Perhaps the simplest thing would be to contact the authors of these papers for followup questions.)
