A not unreasonable thing to do, and probably what you are expected to do, is to expand $\sqrt{x}$ in powers of $x-9$.
The formula for the $n$-th degree Taylor polynomial for $f(x)$ in powers of $x-a$ is
$$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a)^n.$$
In our case we have $f(x)=x^{1/2}$, $a=9$, and $n=2$. To finish getting the second degree polynomial, you need to calculate $f(9)$, $f'(9)$, and $f''(9)$.
We now turn to the error term. By the Lagrange form of the remainder, the absolute value of the error if we use the Taylor polynomial of degree $2$ is
$$\frac{|f'''(\xi)|}{3!}|x-9|^3,$$
where $f(x)=x^{1/2}$ and $\xi$ is a number between $x$ and $9$. In our case, we have $x=8$.
Note that $f'''(x)=\frac{3}{8}x^{-5/2}$. Between $8$ and $9$, it is positive and bounded above by $\frac{3}{8}8^{-5/2}$. Thus, putting things together, and noting that $|8-9|=1$, we find that the absolute value of the error is less than $\frac{3}{(8)(3!)}8^{-5/2}$.
For the sine function, the best thing to do is to note that when $|x|\lt 1$ then the usual series for $\sin x$ is an alternating series. For such a series, the absolute value of the error is less than the absolute value of the first "neglected" term. In our case, that term is $-\frac{x^7}{7!}$, so in the interval $[0,1]$ the error has absolute value $\lt \frac{1}{7!}$.
Alternately, but less attractively, we can use the Lagrange formula for the remainder. It is useful to note that the given series up to $\frac{x^5}{5!}$ can be considered to be the expansion up to the term in $x^6$, which happens to be $0$. Thus the Lagrange formula gives absolute value of the error equal to
$$\frac{|f^{(7)}(\xi)|}{7!}|x|^7,$$
where $f(x)=\sin x$. The $7$-th derivative of $f(x)$ is $-\cos x$, so the absolute value of the $7$-th derivative is $\lt 1$. That gives us that the absolute value of the error if $|x|\lt 1$ is $\lt \frac{1}{7!}$.
The remainder is $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $x$ and $a$. In this case, the derivatives of $f(x)$ are all $\pm\sin x$ or $\pm \cos x$, so $|f^{(n+1)}(c)|\leq 1$. Thus $|R_n|\leq \frac{|x-\frac{\pi}{6}|^{n+1}}{(n+1)!}$. You can plug in $n=4$ to finish the problem.
Edit: The preceding argument was for all of $\mathbb R$, but the problem is for the specific interval $[0,\pi/3]$, and the OP wants an absolute numerical bound. Restricting to $[0,\pi/3]$ does not let us improve our bound on the fifth derivative $f^{(5)}(x)=\cos(x)$ since $\cos(x)$ achieves the value $1$ on that interval. So
$$R_4\leq \frac{1}{5!}|x-\pi/6|^5.$$
To get an absolute numerical bound, notice that $|x-\pi/6|\leq \pi/6$ on the given interval.
Best Answer
Taylor series for the function can be written as $$\ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}, x\in(-1,1].$$
(a) The fourth Taylor ploynomial at $x=1$ is $(-1)^{n+1}\frac{x^n}{n}|_{x=1}=-\frac{x^4}{4}$.
(b) We cannot approximate the value of $\ln(1+1.2)$ since $x\in(-1,1]$.
Hence, I guess you mean Maclaurin series of $\ln(1+x)$ at $x_{0} = 1$. Maclaurin series of $\ln(1+x)$ at 1 is $$\ln(1+x) = \ln(2) + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2^nn}(x-1)^n, x\in\mathbf{R}$$ (a) The fourth Taylor ploynomial at $x=1$ is $\frac{(-1)^{n-1}}{2^nn}(x-1)^n.$
(b) (c) I didn't fully understand the meaning "correct 4 decimal".