Answer: \(\boxed{T=\dfrac{7W}{30}}\), \(\boxed{\lambda=\dfrac{7W}{9}}\), and \(\boxed{80.7^\circ}\).

Since \(\sin\theta=\frac35\), we have

\[\cos\theta=\frac45.\]

We are also given

\[\cos2\theta=\frac7{25}.\]

(i) Take moments about the hinge \(A\). The moment of the tension about \(A\) is

\[T(3a\sin\theta).\]

The weight of the rod acts at its midpoint, so its distance from \(A\) along the rod is \(1.5a\). The perpendicular distance for the weight is \(1.5a\cos2\theta\).

Hence

\[T(3a\sin\theta)=W(1.5a\cos2\theta).\]

Substitute \(\sin\theta=\frac35\) and \(\cos2\theta=\frac7{25}\):

\[T\left(3a\cdot\frac35\right)=W\left(1.5a\cdot\frac7{25}\right).\]

Solving gives

\[\boxed{T=\frac{7W}{30}}.\]

(ii) Find the length of the string. The horizontal distance from the wall to \(B\) is

\[5a-3a\cos2\theta.\]

Since the string makes angle \(\theta\) with the horizontal,

\[OB\cos\theta=5a-3a\cos2\theta.\]

Therefore

\[OB=\frac{5a-3a(7/25)}{4/5}=\frac{26a}{5}.\]

The natural length of the string is \(4a\), so the extension is

\[\frac{26a}{5}-4a=\frac{6a}{5}.\]

By Hooke's law,

\[T=\frac{\lambda x}{4a},\]

where \(x=\frac{6a}{5}\). Thus

\[\frac{7W}{30}=\frac{\lambda(6a/5)}{4a}=\frac{3\lambda}{10}.\]

Hence

\[\boxed{\lambda=\frac{7W}{9}}.\]

(iii) Let the force at the hinge have horizontal component \(X\) and vertical component \(Y\).

The horizontal component balances the horizontal component of the tension:

\[X=T\cos\theta=\frac{7W}{30}\cdot\frac45=\frac{14W}{75}.\]

Vertically, the hinge must balance the weight together with the vertical component of the tension, so

\[Y=W+T\sin\theta=W+\frac{7W}{30}\cdot\frac35=\frac{57W}{50}.\]

If the force at \(A\) makes angle \(\phi\) with the horizontal, then

\[\tan\phi=\frac{Y}{X}=\frac{57W/50}{14W/75}=\frac{171}{28}.\]

Therefore

\[\phi=\tan^{-1}\left(\frac{171}{28}\right)=\boxed{80.7^\circ}.\]