(a) To show that \(p\) satisfies \(\tan p = \frac{1}{1+p}\), start by equating the functions at \(x = p\):

\(\cos p = \frac{k}{1+p}\).

Differentiate both functions:

\(\frac{d}{dx}(\cos x) = -\sin x\)

\(\frac{d}{dx}\left(\frac{k}{1+x}\right) = -\frac{k}{(1+x)^2}\).

Equate the derivatives at \(x = p\):

\(-\sin p = -\frac{k}{(1+p)^2}\).

Substitute \(k = (1+p)\cos p\) into the derivative equation:

\(-\sin p = -\frac{(1+p)\cos p}{(1+p)^2}\).

Simplify to obtain:

\(\tan p = \frac{1}{1+p}\).

(b) Use the iterative formula:

\(p_{n+1} = \arctan\left(\frac{1}{1+p_n}\right)\).

Start with an initial guess, say \(p_0 = 0.5\), and iterate:

\(p_1 = \arctan\left(\frac{1}{1+0.5}\right) = 0.54042\)

\(p_2 = \arctan\left(\frac{1}{1+0.54042}\right) = 0.56609\)

\(p_3 = \arctan\left(\frac{1}{1+0.56609}\right) = 0.56777\)

\(p_4 = \arctan\left(\frac{1}{1+0.56777}\right) = 0.56799\)

\(p_5 = \arctan\left(\frac{1}{1+0.56799}\right) = 0.56800\)

The value of \(p\) correct to 3 decimal places is \(0.568\).

(c) Substitute \(p = 0.568\) back into \(\cos p = \frac{k}{1+p}\):

\(\cos 0.568 = \frac{k}{1+0.568}\).

Calculate \(\cos 0.568 \approx 0.842\).

\(0.842 = \frac{k}{1.568}\).

\(k = 0.842 \times 1.568 \approx 1.32\).