If two similar springs each of spring constant $$K_1$$ are joined in series, the new spring constant and time period would be changed by a factor:

When two identical springs, each of spring constant $$K_1$$, are connected in series, the equivalent spring constant is given by $$\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_1} = \frac{2}{K_1}$$, which gives $$K_{eq} = \frac{K_1}{2}$$.

So the new spring constant is $$\frac{1}{2}$$ times the original spring constant $$K_1$$.

The time period of oscillation for a spring-mass system is $$T = 2\pi\sqrt{\frac{m}{K}}$$. The ratio of the new time period to the original time period is $$\frac{T_{new}}{T_{old}} = \sqrt{\frac{K_1}{K_{eq}}} = \sqrt{\frac{K_1}{K_1/2}} = \sqrt{2}$$.

Therefore the time period changes by a factor of $$\sqrt{2}$$.

Hence the correct answer is Option 4: $$\frac{1}{2}, \sqrt{2}$$.