Find the value of k: limx->1 (x4 - 1)/(x − 1) = limx->k (x3 - k3)/(x2 - k2)

The LHS is in the form $\lim_{x \to c} \frac{x^n - c^n}{x - c} = n c^{n-1}$. Here, $c=1$ and $n=4$.
$$
\lim_{x \to 1} \frac{x^4 - 1^4}{x - 1} = 4 \cdot (1)^{4-1} = 4 \cdot 1^3 = 4
$$

The RHS is an indeterminate form $\frac{0}{0}$ when $x=k$. We can rewrite it by dividing the numerator and denominator by $(x-k)$:
$$
\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{\frac{x^3 - k^3}{x - k}}{\frac{x^2 - k^2}{x - k}}
$$
Apply the limit formula $\lim_{x \to c} \frac{x^n - c^n}{x - c} = n c^{n-1}$ to both the numerator and the denominator.  
Numerator: $\lim_{x \to k} \frac{x^3 - k^3}{x - k} = 3k^{3-1} = 3k^2$.  
Denominator: $\lim_{x \to k} \frac{x^2 - k^2}{x - k} = 2k^{2-1} = 2k$.  
So, the RHS becomes:
$$
\frac{3k^2}{2k}
$$
Assuming $k \ne 0$, we can simplify this to $\frac{3k}{2}$. If $k=0$, the RHS would be $\lim_{x \to 0} \frac{x^3}{x^2} = \lim_{x \to 0} x = 0$, which is not equal to the LHS (4). Thus, $k \ne 0$.

$$
4 = \frac{3k}{2}
$$
$$
8 = 3k
$$
$$
k = \frac{8}{3}
$$
The value of $k$ is $\boxed{\frac{8}{3}}$.

The limit is an indeterminate form $\frac{0}{0}$ when $x=-a$. We use the formula $\lim_{x \to c} \frac{x^n - c^n}{x - c} = n c^{n-1}$.  
Since $n=9$ is an odd number, $a^9 = -(-a)^9$.  
So, $x^9 + a^9 = x^9 - (-a)^9$.  
Also, $x+a = x - (-a)$.  
Therefore, the limit can be written as:
$$
\lim_{x \to -a} \frac{x^9 - (-a)^9}{x - (-a)}
$$

Here, $c = -a$ and $n = 9$.  
The limit is $9 \cdot (-a)^{9-1} = 9 \cdot (-a)^8$.  
Since any non-zero number raised to an even power is positive, $(-a)^8 = a^8$.  
So, the limit evaluates to $9a^8$.

$$
9a^8 = 9
$$
$$
a^8 = 1
$$
To find $a$, we take the eighth root of both sides. Since the exponent is even, there will be two real solutions (positive and negative).
$$
a = \pm \sqrt[8]{1}
$$
$$
a = \pm 1
$$
The possible values of $a$ are $\boxed{\pm 1}$.