The variables $x$ and $y$ are related by the differential equation $\pi \dfrac{dy}{dx} + y(3 - \pi \tan x) = 0$.

1. Differentiate the given substitution with respect to $x$.

Using the product rule on $y = z\sec x$:
$\frac{dy}{dx} = \frac{dz}{dx}\sec x + z\sec x \tan x$

2. Substitute $y$ and $\frac{dy}{dx}$ into the given differential equation.

Substitute into $\pi \frac{dy}{dx} + y(3 - \pi \tan x) = 0$:
$\pi \left( \frac{dz}{dx}\sec x + z\sec x \tan x \right) + (z\sec x)(3 - \pi \tan x) = 0$

3. Expand and simplify to obtain the given expression for $\frac{dz}{dx}$.

$\pi \sec x \frac{dz}{dx} + \pi z\sec x \tan x + 3z\sec x - \pi z\sec x \tan x = 0$
$\pi \sec x \frac{dz}{dx} + 3z\sec x = 0$
Dividing by $\pi \sec x$ (since $\sec x \neq 0$):
$\frac{dz}{dx} + \frac{3z}{\pi} = 0$
$\frac{dz}{dx} = -\frac{3z}{\pi}$

4. Separate the variables and integrate.

$\int \frac{1}{z} \,dz = \int -\frac{3}{\pi} \,dx$
$\ln z = -\frac{3}{\pi}x + C$

5. Substitute back $z = y\cos x$ and apply the initial conditions to find the constant $C$.

Since $y = z\sec x$, we have $z = y\cos x$.
$\ln (y\cos x) = -\frac{3}{\pi}x + C$
Substitute $x = \frac{\pi}{3}$ and $y = 2$:
$\ln \left(2\cos\frac{\pi}{3}\right) = -\frac{3}{\pi}\left(\frac{\pi}{3}\right) + C$
$\ln \left(2 \cdot \frac{1}{2}\right) = -1 + C$
$\ln 1 = -1 + C \implies C = 1$

6. Express the solution in the required exponential form and identify constants $a$ and $b$.

Substitute $C = 1$ back into the equation:
$\ln (y\cos x) = 1 - \frac{3}{\pi}x$
$y\cos x = e^{1 - \frac{3}{\pi}x}$
Comparing with $y\cos x = e^{a+bx}$, we have:
$a = 1, \; b = -\frac{3}{\pi}$

1. Identify the condition for vertical asymptotes from the solution's equation.

From part (a), the solution can be rewritten as $y = \frac{e^{a+bx}}{\cos x}$.

Vertical asymptotes occur where the denominator approaches zero while the numerator remains non-zero. Since $e^{a+bx} > 0$ for all real $x$, the vertical asymptotes correspond exactly to the roots of:
$\cos x = 0$

2. Solve for the two asymptotes closest to the y-axis.

Solving $\cos x = 0$ gives:
$x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, \dots$

The two vertical asymptotes with equations closest to the $y$-axis ($x=0$) are:
$x = \frac{\pi}{2} \quad \text{and} \quad x = -\frac{\pi}{2}$