There is an easy tutorial on Thorlabs web page explaining this. It's just a single application of Snell's law assuming you know the index of the fiber and the cladding. The numerical aperture is defined by the the maximum half-acceptance angle $\theta_a$ associated with total internal reflection.
Let $n_i$ be the index of the outside material (usually 1) and $\theta_c$ be the critical angle for total internal reflection. Then
$$\sin \theta_c = n_c/n_f = \cos \theta_t = \sqrt{1- \sin^2 \theta _t}$$
So the numerical aperture $NA$, is
$$ NA = n_i \sin \theta_a = \sqrt{n_f ^2 - n_c ^2}$$
If $n_i$ is 1 and $\theta_a$ is small, then you just the NA is approximately $\theta_a$.
Here is the link to Thorlabs: https://www.thorlabs.com/tutorials.cfm?tabID=17485628-68dd-4d22-ad17-6dd4520974c7
You need to differentiate between the components of a vector and the vector itself.
The $n$ coordinate vectors $\partial_j$ constitute a basis of the n-dimensional tangent space $T_p(M)$ at each point $p\in M$. Sometimes these coordinate bases induced by the charts $(U,h)$ are referred to as the natural or canonical bases.
Relative to a given chart (U,h) with coordinate functions $x^1,\dots, x^n$ any vector $v\in T_p(M)$ admits the representation
$$v=v^i\frac{\partial }{\partial x^i}$$
When we say that $v^i$ is a contravariant vector we actually mean that the components $v^i$ transforms like a contravariant vector at $p$ (and that $v\in T_p(M)$). The coefficients are uniquely defined by
$$v^i=vx^i$$
For two overlapping charts the vector $v$ is represented by $v=v^i\frac{\partial}{\partial x^i}$ and $v=\bar{v}^j\frac{\partial}{\partial {\bar{x}}^j}$ respectively. With $v^i=v x^i$ and $\bar{v}^j=v\bar{x}^j$ we can make two important observations
$$\bar{v}^j=v^{i}\frac{\partial \bar{x}^j}{\partial x^i}\tag{1}$$
$$v=\bar{v}^j\frac{\partial }{\partial \bar{x}^j}=v^h\frac{\partial\bar{x}^j}{\partial x^h}\frac{\partial}{\partial \bar{x}^j}=v^h\frac{\partial}{\partial x^h}\tag{2}$$
Now $(1)$ tells us that indeed the components $v^i$ transforms like the components of a contravariant vector. Further, since $(2)$ is valid for any arbitrary $v^i$ we conclude that
$$\frac{\partial }{\partial x^h}=\frac{\partial \bar{x}^j}{\partial x^h}\frac{\partial }{\partial \bar{x}^j}$$
So the basis elements for the (contravariant) tangent space itself transform like covariant vectors!
I will not go into details on how to construct the dual basis $\{dx^k,\,k=1\dots n\}$ of $T^*_p(M)$ but it follows naturally from the definition of the unique element $df\in T^*_p(M)$ such that $\langle d f,v\rangle=vf$ that any $\omega\in T^*_p(M)$ can be expressed as
$$\omega=\omega_jdx^j$$
Notice again that $\omega_j$ are the components of a covariant tensor, while $dx^j$ are the basis (which transforms contravariantly). The dual tangent space $T^*_p(M)$ is called the cotangent space, and its elements $\omega$ are referred to as covectors or 1-forms.
Specifically with $df=f_hdx^h$ we have $df=\partial_jfdx^j$ (since $\langle df,\partial_j\rangle=f_j$) which is consistent with the customary expression for the differential of a function.
Best Answer
As long as this is not homework:
1.1: Yes.
1.2: If you are taking a class, then you had better learn this equation well, as it is Snell's Law, or the law of refraction. $\alpha_1$ is indeed the angle of incidence from the surface normal. $\alpha_2$ is the angle of refraction from the surface normal. In this case, $\alpha_1 = 45^{\circ}$. You find $\alpha_2$ given $n_1$ and $n_2$:
$$\sin{\alpha_2} = \frac{n_1}{n_2} \sin{\alpha_1} $$
Note that, if $n_1 \le n_2$, then there will always be light refracted, no matter what the angle of incidence is. However, if $n_1 \gt n_2$, then there will be some critical angle $\alpha_c$ at which $\sin{\alpha_2} =1$, beyond which the light does not refract into the surrounding medium but instead reflects back.
In the case pictured, the light reflects back at an incidence of $\alpha_1=45^{\circ}$, $n_2=1$. Thus, the smallest value $n_1$ can have is whatever makes $\sin{\alpha_2}=1$, or $n_1 = 1/\sin{\alpha_1} = \sqrt{2} \approx 1.414$.
1.3: You can get the minimum value of $n_1$ the same way I did in 1.2 above for the fluid at $n_2=1.15$. The maximum value of $n_1$ results using the same condition at $n_2=1.33$.