Given:

1. The angle of projection $$\theta = 30^{\circ}$$.
2. The projectile is at the same height at time $$t_1 = 3 \mathrm{~s}$$ and $$t_2 = 5 \mathrm{~s}$$.
3. The acceleration due to gravity $$g = 10 \mathrm{~m/s^2}$$.

We need to find the initial speed of projection, $$u$$.

We can use the following equation to find the vertical displacement, $$y$$, at any time $$t$$:

$$y = u_yt - \frac{1}{2}gt^2$$

Where $$u_y$$ is the initial vertical component of the velocity, $$u_y = u \sin \theta$$.

Since the projectile is at the same height at $$t_1$$ and $$t_2$$, we can write:

$$u_yt_1 - \frac{1}{2}gt_1^2 = u_yt_2 - \frac{1}{2}gt_2^2$$

Substitute the values of $$t_1$$ and $$t_2$$:

$$u_y(3) - \frac{1}{2}(10)(3)^2 = u_y(5) - \frac{1}{2}(10)(5)^2$$

Now, let's find the initial vertical component of the velocity, $$u_y$$:

$$u_y = u \sin \theta = u \sin(30^{\circ}) = \frac{1}{2}u$$

Substitute $$u_y$$ in the equation:

$$\frac{1}{2}u(3) - \frac{1}{2}(10)(3)^2 = \frac{1}{2}u(5) - \frac{1}{2}(10)(5)^2$$

Now, simplify and solve for $$u$$:

$$3u - 90 = 5u - 250$$

$$2u = 160$$

$$u = 80 \mathrm{~m/s}$$

The initial speed of projection is $$80 \mathrm{~m/s}$$.