Given $\lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^2}$
It forms $\frac{0}{0}$ indeterminate form, then apply L'Hospital rules.
$$
\begin{aligned}
\Rightarrow & \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4} \\
& \lim _{x \rightarrow 0} \frac{2 \tan \left(\pi \sec ^4 x\right) \cdot \sec ^2\left(\pi \sec ^4 x\right) \cdot 4 \pi \sec ^4 x \tan x}{4 \pi^2 x^3} \\
\Rightarrow & \lim _{x \rightarrow 0}\left(\frac{\left.2 \tan \left(\pi \sec ^4 x\right) \cdot \sec ^2\left(\pi \sec ^4 x\right) \cdot \sec ^4 x\right)}{\pi x^2}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right) \\
\Rightarrow & \lim _{x \rightarrow 0}\left(\frac{2 \tan \left(\pi \sec ^4 x\right)}{x^2}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\sec ^2\left(\pi \sec ^4 x\right) \cdot \sec ^4 x}{\pi}\right) \cdot 1
\end{aligned}
$$
Again Apply L'Hospital rule,
$$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0}\left(\frac{2 \sec ^2\left(\pi \sec ^4 x\right) \cdot 4 \pi \sec ^4 x \tan x}{2 x}\right) \frac{1}{\pi}\left(\sec ^2\left(\pi \sec ^4 0\right) \cdot \sec ^4(0)\right) \\
& \Rightarrow \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right) \cdot \lim _{x \rightarrow 0} \frac{\left(\sec ^2\left(\pi \sec ^4 x\right) \cdot 4 x \sec ^4 x\right)}{\pi} \cdot\left(\sec ^2(\pi) \cdot 1\right) \\
& \quad \Rightarrow 1 \cdot \sec ^2\left(\pi \sec ^4(0)\right) \cdot 4 \cdot \sec ^4(0) \cdot(-1)^2 \\
& \quad \Rightarrow 1 \cdot(-1)^2 \cdot 4 \cdot(1)=4
\end{aligned}
$$