Solution: 1. We need to prove that the set $\mathcal{B} $ is linearly independent, i.e., if \begin{equation}\label{eq:30Jan2023-1} a \sin x + b\cos x + c \tan x + d \sec x = 0, \end{equation} then $a=b=c=d=0$.

---

Since \eqref{eq:30Jan2023-1} is true for all $x\in \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) $, so we take $x=0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3}$. We will get the following system of linear equations. \begin{equation}\label{eq:30Jan2023-2} \begin{gathered} x= 0 \implies 0\cdot a + 1 \cdot b + 0\cdot c + 1 \cdot d = 0 \\ x= \frac{\pi}{6} \implies \frac{1}{2}\cdot a + \frac{\sqrt{3} }{2} \cdot b + \frac{1}{\sqrt{3} }\cdot c + \frac{2}{\sqrt{3} } \cdot d = 0 \\ x= \frac{\pi}{4} \implies \frac{1}{\sqrt{2} }\cdot a + \frac{1}{\sqrt{2} } \cdot b + 1\cdot c + \sqrt{2} \cdot d = 0 \\ x= \frac{\pi}{3} \implies \frac{\sqrt{3} }{2}\cdot a + \frac{1}{2} \cdot b + \sqrt{3} \cdot c + 2 \cdot d = 0. \end{gathered} \end{equation} The above system will have trivial solution if and only if the determinant of the coefficient matrix will be zero.

---

Note that \begin{align*} \begin{vmatrix} 0 & 1 & 0 & 1 \\[1ex] \frac{1}{2} & \frac{\sqrt{3} }{2} & \frac{1}{\sqrt{3} } & \frac{2}{\sqrt{3} } \\[1ex] \frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} } & 1 & \sqrt{2} \\[1ex] \frac{\sqrt{3} }{2} & \frac{1}{2} & \sqrt{3} & 2 \\ \end{vmatrix} = 0. \end{align*} Thus, the system \eqref{eq:30Jan2023-2} has only trivial solution, and this proves that $a=b=c=d=0$. Therefore, the $\mathcal{B} $ is linearly independent.

---

This can be solved in another way also. Consider the equation \eqref{eq:30Jan2023-1}. For every $x\in \left( -\frac{\pi}{2},\frac{\pi}{2} \right) $ we have, \begin{align*} & a \sin x + b\cos x + c \tan x + d \sec x = 0\\ \implies & a \sin x\cos x + b\cos^2 x + c \sin x + d = 0\\ \implies & \sin x (a \cos x+c) = - \left( b \cos ^2x + d \right) \\ \implies & \left[ \sin x (a \cos x+c) \right]^2 = \left( b \cos ^2x + d \right)^2 \\ \implies & \sin ^2x \left( a^2 \cos ^2x + 2ac \cos x + c^2 \right) = b^2 \cos ^4x+2bd \cos ^2x + d^2 \\ \implies & (1-\cos ^2x) \left( a^2 \cos ^2x + 2ac \cos x + c^2 \right) \\ & \kern 2cm = b^2 \cos ^4x+2bd \cos ^2x + d^2 \\ \implies & \left( a^2+b^2 \right) \cos ^4x + 2ac \cos ^3x + \left( c^2 -a^2 +2bd \right) \cos^2x \\ & \kern 2cm - 2ac \cos x + (d^2 - c^2) = 0. \end{align*}

---

Note that the above is a polynomial in $\cos x$, and it is zero for all $x\in \left( -\frac{\pi}{2},\frac{\pi}{2} \right) $, so the polynomial must be identically zero. Hence, \[ a^2+b^{2} =0 ,~2ac=0,~ c^2 -a^{2} +2bd = 0,~2ac = 0,~d^2 - c^2 =0. \] The above proves that $a=b=c=d=0$.

---

2. Note that \begin{align*} \sin ^2x+\cos ^2x = 1 \text{ and } \sec ^2x -\tan ^2x = 1. \end{align*} So we have, \begin{align*} & \left( \sin ^2x+\cos ^2x \right) + \left( \tan ^2x-\sec ^2x \right) = 0 \\ \implies & \left( \sin ^2x+\cos ^2x \right) + \left( \tan ^2x-\sec ^2x \right) = 0 \\ \implies & T (\sin x) + T(\cos x) + T(\tan x) - T(\sec x) = 0\\ \implies & T (\sin x+\cos x+\tan x-\sec x) = 0. \end{align*} Thus, \[ \sin x+\cos x+\tan x-\sec x \in \ker T. \]

---

Now note that, $\dim W = 4$ and \begin{align*} \sin^2x,\cos ^2x,\tan ^2x \in \mathrm{Im}(T). \end{align*} We also observe that above are linearly independent. \begin{align*} & a \sin ^2x + b \cos ^2x + c \tan ^2x = 0 \\ \implies & 0\cdot a + 1\cdot b + 0 \cdot c = 0 \\ \implies & \frac{1}{2}\cdot a + \frac{1}{2}\cdot b + 1 \cdot c = 0 \\ \implies & \frac{3}{4}\cdot a + \frac{1}{4}\cdot b + 3 \cdot c = 0. \end{align*} The coefficient matrix has zero determinant and hence, the system has trivial solution. Therefore, $\left\{ \sin ^2x,\cos ^2x,\tan ^2x \right\} $ is linearly independent.

---

Now using the rank-nullity theorem we have \[ \mathrm{rank} + \mathrm{nullity} = \dim W = 4. \] Since, $\mathrm{rank} \ge 3 $ and $\mathrm{nullity} \ge 1 $, so these two inequalities must be equality. Therefore, \[ \mathrm{rank} = 3,\text{ and } \mathrm{nullity} =1. \] So we proved that, \[ \ker T = \mathrm{span} \left( \sin x+\cos x+\tan x-\sec x\ \right). \] So, a basis of kernel will be \[ \textcolor{blue}{\boxed{\text{basis of kernel } = \{\sin x+\cos x+\tan x-\sec x\}}} \]