Answer: The string breaks when \(\cos\theta=\dfrac35\), the speed then is \(\boxed{\sqrt{\dfrac{8ag}{5}}}\), and later \(\boxed{OP=\dfrac{20a}{9}}\) when \(P\) is vertically below \(O\).

(i) Let the initial speed be

\[u=\sqrt{\frac{2ag}{5}}.\]

When \(OP\) makes angle \(\theta\) with the downward vertical, the particle has fallen a vertical distance \(a\cos\theta\) from its initial horizontal position.

By conservation of energy,

\[\frac12mv^2=\frac12mu^2+mga\cos\theta.\]

So

\[v^2=u^2+2ga\cos\theta.\]

For the radial equation, tension acts towards \(O\). The component of weight away from \(O\) is \(mg\cos\theta\), so

\[T-mg\cos\theta=\frac{mv^2}{a}.\]

Thus

\[T=\frac{mv^2}{a}+mg\cos\theta.\]

Substitute \(v^2=u^2+2ga\cos\theta\):

\[T=\frac{m}{a}\left(u^2+2ga\cos\theta\right)+mg\cos\theta.\]

Therefore

\[T=\frac{mu^2}{a}+3mg\cos\theta.\]

Since \(u^2=\frac{2ag}{5}\),

\[\frac{mu^2}{a}=\frac25mg.\]

The string breaks when \(T=\frac{11}{5}mg\). Hence

\[\frac{11}{5}mg=\frac25mg+3mg\cos\theta.\]

Cancel \(mg\):

\[\frac{11}{5}=\frac25+3\cos\theta.\]

So

\[3\cos\theta=\frac95,\qquad \cos\theta=\frac35,\]

as required.

Now use the energy equation again:

\[v^2=u^2+2ga\cos\theta.\]

Thus

\[v^2=\frac{2ag}{5}+2ga\left(\frac35\right)=\frac{2ag}{5}+\frac{6ag}{5}=\frac{8ag}{5}.\]

So the speed when the string breaks is

\[\boxed{v=\sqrt{\frac{8ag}{5}}}.\]

(ii) Since \(\cos\theta=\frac35\), we have \(\sin\theta=\frac45\).

At the instant the string breaks, the horizontal distance of \(P\) from the vertical line through \(O\) is

\[a\sin\theta=\frac45a.\]

The horizontal component of the velocity towards the vertical line is

\[v\cos\theta=\frac35v.\]

So the time taken to reach the vertical line below \(O\) is

\[t=\frac{(4a/5)}{(3v/5)}=\frac{4a}{3v}.\]

The vertical component of the velocity downwards is

\[v\sin\theta=\frac45v.\]

During this time, the additional vertical distance fallen is

\[h=\left(\frac45v\right)t+\frac12gt^2.\]

Using \(t=\frac{4a}{3v}\), the first term is

\[\frac45v\cdot\frac{4a}{3v}=\frac{16a}{15}.\]

Also, because \(v^2=\frac{8ag}{5}\),

\[t^2=\frac{16a^2}{9v^2}=\frac{16a^2}{9(8ag/5)}=\frac{10a}{9g}.\]

Therefore

\[\frac12gt^2=\frac12g\cdot\frac{10a}{9g}=\frac{5a}{9}.\]

Hence

\[h=\frac{16a}{15}+\frac{5a}{9}=\frac{73a}{45}.\]

At the breaking point, the particle is already \(a\cos\theta=\frac35a\) below \(O\). Therefore when it is vertically below \(O\),

\[OP=\frac35a+h=\frac{3a}{5}+\frac{73a}{45}=\frac{100a}{45}=\boxed{\frac{20a}{9}}.\]