#### Answer

\begin{align*}
T_{2}&=1+\frac{2}{1 !}\left(x-\frac{\pi}{4}\right)+\frac{4}{2 !}\left(x-\frac{\pi}{4}\right)^{2}\\
T_{3}&=1+\frac{2}{1 !}\left(x-\frac{\pi}{4}\right)+\frac{4}{2 !}\left(x-\frac{\pi}{4}\right)^{2}+\frac{16}{3 !}\left(x-\frac{\pi}{4}\right)^{3}
\end{align*}

#### Work Step by Step

Since
\begin{array}{ccc}
{\tan \left(\frac{\pi}{4}\right)} & {=} & {1} \\
{\frac{d}{d x}(\tan (x))\left(\frac{\pi}{4}\right)} & {=} & {2} \\
{\frac{d^{2}}{d x^{2}}(\tan (x))\left(\frac{\pi}{4}\right)} & {=} & {4} \\
{\frac{d^{3}}{dx^{3}}(\tan (x))\left(\frac{\pi}{4}\right)} & {=} & {16}
\end{array}
Then
\begin{align*}
T_{2}&=1+\frac{2}{1 !}\left(x-\frac{\pi}{4}\right)+\frac{4}{2 !}\left(x-\frac{\pi}{4}\right)^{2}\\
T_{3}&=1+\frac{2}{1 !}\left(x-\frac{\pi}{4}\right)+\frac{4}{2 !}\left(x-\frac{\pi}{4}\right)^{2}+\frac{16}{3 !}\left(x-\frac{\pi}{4}\right)^{3}
\end{align*}