Answer
$$T_2=T_3=1+2x^2$$
Work Step by Step
Given $$f(x)=\cosh 2 x, \quad a=0$$
Since
\begin{aligned}
f(x) &=\cosh 2 x&& f(0) &=1\\
f^{\prime}(x) &=2\sinh 2x & & f^{\prime}(0) &=0\\
f^{\prime \prime}(x) &=4\cosh 2x & & f^{\prime \prime}(0) &=4 \\
f^{\prime \prime \prime}(x) &=8\sinh 2x & & f^{\prime \prime \prime}(0) &=0
\end{aligned}
Then
\begin{aligned}
T_{2}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2}(x-a)^{2} \\
&=1+2x^2
\end{aligned}
and
\begin{aligned}
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} \\
&=1+\frac{0}{1}(x-0)+\frac{4}{2!}(x-0)^{2}+\left(\frac{0}{6}\right)(x-0^{3} \\
&=1+2x^2
\end{aligned}
