Solution

Given grammar is

$S\to XY$

$X\to YaY\mid a$

$Y\to bbX$

Since $S\to XY$, the string contains one part from $X$ and one part from $Y$.

Now, $X$ can generate at least one $a$.

Also, $Y\to bbX$, so $Y$ also contains one $X$, which again gives at least one $a$.

Therefore, every string generated by $S$ has at least two $a$'s.

So, statement $(c)$ is true.

Also, every use of $Y$ contributes $bb$, that is $2$ occurrences of $b$.

So, the number of $b$'s is always in multiple of $2$.

Therefore, statement $(d)$ is true.

The grammar contains $bb$, so alternate $a$ and $b$ condition is false.

Also, consecutive three $a$'s are not necessarily produced.