Answer
$$ T_2(x)=x$$
$$T_3(x)= x+\frac{1}{3} x^{3}$$
Work Step by Step
Given $$f(x)=\tan x, \quad a=0$$
Since
\begin{array}{rlrl}
f(x) & =\tan x & f(0)&=0 \\
f^{\prime}(x) & =\sec ^{2} x & f^{\prime}(0) & =1 \\
f^{\prime \prime}(x) & =2 \sec ^{2} x \tan x & f^{\prime \prime}(0) & =0 \\
f^{\prime \prime \prime}(x) & =2 \sec ^{4} x+4 \sec ^{2} x \tan ^{2} x & f^{\prime \prime \prime}(0) & =2
\end{array}
Then
\begin{align*}
T_{2}(x)&=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2}\\
&=0+1(x-0)+\frac{0}{2}(x-0)^{2}=x
\end{align*}
and
\begin{align*}
T_{3}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{6}(x-a)^{3} \\
&=0+1(x-0)+\frac{0}{2}(x-0)^{2}+\frac{2}{6}(x-0)^{3}\\
&=x+\frac{1}{3} x^{3}
\end{align*}
