Given f(x) = x3 − hx2 − 13x + k. If (x-1) is a factor, then f(1) = 0.

Given $f(x) = x^3 - hx^2 - 13x + k$. If $(x-1)$ is a factor, then $f(1) = 0$.

$$f(1) = (1)^3 - h(1)^2 - 13(1) + k = 0$$ $$1 - h - 13 + k = 0$$ $$-h + k - 12 = 0$$ $$k - h = 12 \quad (*)$$

If $(x+3)$ is a factor, then $f(-3) = 0$.

$$f(-3) = (-3)^3 - h(-3)^2 - 13(-3) + k = 0$$ $$-27 - 9h + 39 + k = 0$$ $$-9h + k + 12 = 0$$ $$k - 9h = -12 \quad (**)$$

Subtract equation $(**)$ from equation $(*)$:

$$(k - h) - (k - 9h) = 12 - (-12)$$ $$k - h - k + 9h = 12 + 12$$ $$8h = 24$$ $$h = \frac{24}{8}$$ $$h = 3$$

$$k - 3 = 12$$ $$k = 12 + 3$$ $$k = 15$$

The values are $h=3$ and $k=15$.

$$\boxed{h=3, k=15}$$

23.

Given $p(x) = (x^2 - 3x + 2)(x^2 + 7x + a)$. If $(x+4)$ is a factor of $p(x)$, then $p(-4) = 0$.

$$p(-4) = ((-4)^2 - 3(-4) + 2)((-4)^2 + 7(-4) + a) = 0$$ $$(16 + 12 + 2)(16 - 28 + a) = 0$$ $$(30)(-12 + a) = 0$$

Since $30 \neq 0$, we must have:

$$-12 + a = 0$$ $$a = 12$$

Given $q(x) = (x^2 + 5x + 4)(x^2 - 5x + b)$. If $(x-1)$ is a factor of $q(x)$, then $q(1) = 0$.

$$q(1) = ((1)^2 + 5(1) + 4)((1)^2 - 5(1) + b) = 0$$ $$(1 + 5 + 4)(1 - 5 + b) = 0$$ $$(10)(-4 + b) = 0$$

Since $10 \neq 0$, we must have:

$$-4 + b = 0$$ $$b = 4$$

The values are $a=12$ and $b=4$.

$$\boxed{a=12, b=4}$$