Answer
a) $\color{blue}{c = 5}$
b) 4th moment, $\color{blue}{\mu_4}$
Work Step by Step
a)
$\begin{align*}
1 &= \int_\mathbb{R} f_Y(y)\ dy \\
&= \int_1^\infty c\cdot y^{-6}\ dy \\
&= c\cdot \frac{y^{-5}}{-5}\ \Biggr\vert_1^\infty \\
&= -\frac{c}{5}\cdot\left( \infty^{-5} - 1^{-5} \right) \\
&= -\frac{c}{5}\cdot\left( 0 - 1\right) \\
1 &= \frac{c}{5} \\
\color{blue}{c}\ &\color{blue}{= 5}
\end{align*}$
b)
$\begin{align*}
\mu_r &= E(Y^r),\ r=1,2,3,\ldots \\
&= \int_\mathbb{R} y^r\cdot f_Y(y)\ dy \\
&= \int_1^\infty y^r\cdot 5y^{-6}\ dy \qquad [\ \text{since}\ f_Y(y) = 5\cdot y^{-6},\ y\gt 1\ ] \\
&= \int_1^\infty 5y^{r-6}\ dy \\
&= \begin{cases}
5\cdot \dfrac{y^{r-5}}{r-5}\; \biggr\vert_1^\infty, & r=1,2,3,4,6,7,8, \ldots \\
5\ln y\; \biggr\vert_1^\infty, & r = 5
\end{cases} \\ \\
&= \begin{cases}
\dfrac{5}{r-5}\left( \infty^{r-5} - 1^{r-5} \right), & r=1,2,3,4,6,7,8,\ldots \\
5(\ln \infty - \ln 1), & r =5
\end{cases} \\ \\
&= \begin{cases} \dfrac{5}{r-5}\left( \dfrac{1}{\infty^{5-r}} - 1 \right), & r=1,2,3,4 \\
5(\ln \infty - \ln 1), & r =5 \\
\dfrac{5}{r-5}\left( \infty^{r-5} - 1^{r-5} \right), & r=6,7,8,\ldots
\end{cases} \\ \\
&= \begin{cases} \dfrac{5}{r-5}\left( \dfrac{1}{\infty} - 1 \right), & r=1,2,3,4 \\
5(\infty -0), & r =5 \\
\dfrac{5}{r-5}\left( \infty - 1 \right), & r=6,7,8,\ldots
\end{cases} \\ \\
&= \begin{cases} \dfrac{5}{r-5}\left(0 - 1 \right), & r=1,2,3,4 \\
5\cdot \infty, & r =5 \\
\dfrac{5}{r-5}\cdot \infty, & r=6,7,8,\ldots
\end{cases} \\ \\
&= \begin{cases} \dfrac{5}{5-r}, & r=1,2,3,4 \\
\infty, & r =5,6,7,8, \ldots
\end{cases} \\ \\
\color{blue}{\mu_r}\ &\color{blue}{= \dfrac{5}{5-r},\ r=1,2,3,4.}
\end{align*}$
Thus, the highest moment of $Y$ that exists is the 4th moment, $\color{blue}{\mu_4}$.
