To solve these problems, we can use trigonometric identities and properties. Here are the calculations:

a) \( A = \frac{\sin 32^\circ}{\cos 58^\circ} \)

Since \(\sin x = \cos (90^\circ - x)\), this becomes:

\[
A = \frac{\sin 32^\circ}{\sin 32^\circ} = 1
\]

b) \( B = \tan 76^\circ - \cot 14^\circ \)

\(\tan 76^\circ = \cot 14^\circ\) (because \(\tan x = \cot (90^\circ - x)\)):

\[
B = \tan 76^\circ - \tan 76^\circ = 0
\]

c) \( A = \frac{\sin 25^\circ}{\cos 65^\circ} \)

Using the same identity as in part a):

\[
A = \frac{\sin 25^\circ}{\sin 25^\circ} = 1
\]

d) \( B = \tan 58^\circ - \cot 32^\circ \)

\(\tan 58^\circ = \cot 32^\circ\):

\[
B = \tan 58^\circ - \tan 58^\circ = 0
\]

e) \( A = \sin 30^\circ + \sin 60^\circ \)

Knowing \(\sin 30^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):

\[
A = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}
\]

f) \( B = \cos 60^\circ + \cos 30^\circ \)

Knowing \(\cos 60^\circ = \frac{1}{2}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):

\[
B = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}
\]

g) \( B = \sin 30^\circ + \sin 60^\circ + \cos 45^\circ \)

Knowing \(\sin 30^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), and \(\cos 45^\circ = \frac{\sqrt{2}}{2}\):

\[
B = \frac{1}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} = \frac{1 + \sqrt{3} + \sqrt{2}}{2}
\]

These are the simplified results for each of the expressions.