To find the maximum values of the expressions given, we can analyze each one separately:

a) \( f(x) = -x^2 + 3x + 5 \)

This is a quadratic function in the form \( ax^2 + bx + c \), where \( a = -1 \), \( b = 3 \), and \( c = 5 \). Since \( a < 0 \), the parabola opens downwards, and the maximum value occurs at the vertex.

The vertex \( x \)-coordinate is given by \( x = -\frac{b}{2a} = -\frac{3}{2(-1)} = \frac{3}{2} \).

Plug this back into the function to find the maximum value:
\[ f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 5 = \frac{9}{4} + \frac{9}{2} + 5 = \frac{29}{4} \]

b) \( g(x) = -4y^2 + 10y + 10 \)

This is also a quadratic function in \( y \), with \( a = -4 \), \( b = 10 \), \( c = 10 \).

The vertex \( y \)-coordinate is given by \( y = -\frac{b}{2a} = -\frac{10}{2(-4)} = \frac{5}{4} \).

Plug this back into the function to find the maximum value:
\[ g\left(\frac{5}{4}\right) = -4\left(\frac{5}{4}\right)^2 + 10\left(\frac{5}{4}\right) + 10 = -\frac{25}{4} + \frac{50}{4} + \frac{40}{4} = \frac{65}{4} \]

c) \( h(x, y) = -x^2 - 4y^2 + 2x + 10y - 8 \)

This function depends on two variables, and it is a downward-opening paraboloid. To find the maximum, we set the partial derivatives with respect to \( x \) and \( y \) equal to zero:

For \( x \):
\[ \frac{\partial h}{\partial x} = -2x + 2 = 0 \Rightarrow x = 1 \]

For \( y \):
\[ \frac{\partial h}{\partial y} = -8y + 10 = 0 \Rightarrow y = \frac{5}{4} \]

Substitute \( x = 1 \) and \( y = \frac{5}{4} \) back into the function:
\[ h(1, \frac{5}{4}) = -(1)^2 - 4\left(\frac{5}{4}\right)^2 + 2(1) + 10\left(\frac{5}{4}\right) - 8 = -1 - \frac{25}{4} + 2 + \frac{50}{4} - 8 = \frac{-33}{4} + \frac{75}{4} - 8 \]
\[ = \frac{42}{4} - 8 = \frac{42}{4} - \frac{32}{4} = \frac{10}{4} = \frac{5}{2} \]

d) \( k(x, y) = x^2 - 4y^2 + 2xy + 2x + 10y - 8 \)

Finding the maximum of this function involves setting the partial derivatives to zero:

For \( x \):
\[ \frac{\partial k}{\partial x} = 2x + 2y + 2 = 0 \]

For \( y \):
\[ \frac{\partial k}{\partial y} = -8y + 2x + 10 = 0 \]

Solving these simultaneous equations:
\[ 2x + 2y + 2 = 0 \]
\[ -8y + 2x + 10 = 0 \]

Solving these, we find:
\[ x = -1 - y \]
Substitute into the second equation:
\[ -8y + 2(-1-y) + 10 = 0 \]
\[ -8y - 2 - 2y + 10 = 0 \]
\[ -10y + 8 = 0 \]
\[ y = \frac{4}{5} \]

Now \( x = -1 - \frac{4}{5} = -\frac{9}{5} \)

Substitute \( x = -\frac{9}{5} \) and \( y = \frac{4}{5} \) back into the function:
\[ k\left(-\frac{9}{5}, \frac{4}{5}\right) = \left(-\frac{9}{5}\right)^2 - 4\left(\frac{4}{5}\right)^2 + 2\left(-\frac{9}{5}\right)\left(\frac{4}{5}\right) + 2\left(-\frac{9}{5}\right) + 10\left(\frac{4}{5}\right) - 8 \]

Simplify this to find the maximum value.