Solution:

(i) Given that \[BD\] is a diameter of the circle.

And, the angle in a semicircle is a right angle.

So, \[\angle BCD\text{ }=\text{ }90{}^\circ \]

Also given that,

\[\angle DBC\text{ }=\text{ }58{}^\circ \]

In \[\vartriangle BDC\]

\[\angle DBC\text{ }+\angle BCD\text{ }+\angle BDC\text{ }=\text{ }{{180}^{o}}\]

\[58{}^\circ \text{ }+\text{ }90{}^\circ \text{ }+\angle BDC\text{ }=\text{ }{{180}^{o}}\]

Or,

\[{{148}^{o}}~+\angle BDC\text{ }=\text{ }{{180}^{o}}\]

\[\angle BDC\text{ }=\text{ }{{180}^{o}}-\text{ }{{148}^{o}}\]

Thus, \[\angle BDC\text{ }=\text{ }{{32}^{o}}\]

(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.

So, in cyclic quadrilateral \[BECD\]

\[\angle BEC\text{ }+\angle BDC\text{ }=\text{ }{{180}^{o}}\]

\[\angle BEC\text{ }+\text{ }{{32}^{o}}~=\text{ }{{180}^{o}}\]

So,

\[\angle BEC\text{ }=\text{ }{{148}^{o}}\]