We make a few comments only.
$1.$ Note that $2\pi$ is a period of $\sin x$, or, equivalently, $1$ is a period of $\sin(2\pi x)$.
But $\sin x$ has many other periods, such as $4\pi$, $6\pi$, and so on. However, $\sin x$ has no (positive) period shorter than $2\pi$.
$2.$ If $p$ is a period of $f(x)$, and $H$ is any function, then $p$ is a period of $H(f(x))$. So in particular, $2\pi$ is a period of $\sin^2 x$. However, $\sin^2 x$ has a period which is smaller than $2\pi$, namely $\pi$. Note that $\sin(x+\pi)=-\sin x$, so $\sin^2(x+\pi)=\sin^2 x$. It turns out that $\pi$ is the shortest period of $\sin^2 x$.
$3.$ For sums and products, the general situation is complicated. Let $p$ be a period of $f(x)$ and let $q$ be a period of $g(x)$. Suppose that there are positive integers $a$ and $b$ such that $ap=bq=r$. Then $r$ is a period of $f(x)+g(x)$, and also of $f(x)g(x)$.
So for example, if $f(x)$ has $5\pi$ as a period, and $g(x)$ has $7\pi$ as a period, then $f(x)+g(x)$ and $f(x)g(x)$ each have $35\pi$ as a period. However, even if $5\pi$ is the shortest period of $f(x)$ and $7\pi$ is the shortest period of $g(x)$, the number $35\pi$ need not be the shortest period of $f(x)+g(x)$ or $f(x)g(x)$.
We already had an example of this phenomenon: the shortest period of $\sin x$ is $2\pi$, while the shortest period of $(\sin x)(\sin x)$ is $\pi$. Here is a more dramatic example. Let $f(x)=\sin x$, and $g(x)=-\sin x$. Each function has smallest period $2\pi$. But their sum is the $0$-function, which has every positive number $p$ as a period!
$4.$ If $p$ and $q$ are periods of $f(x)$ and $g(x)$ respectively, then any common multiple of $p$ and $q$ is a period of $H(f(x), g(x))$ for any function $H(u,v)$, in particular when $H$ is addition and when $H$ is multiplication. So the least common multiple of $p$ and $q$, if it exists, is a period of $H(f(x),g(x))$. However, it need not be the smallest period.
$5.$ Periods can exhibit quite strange behaviour. For example, let $f(x)=1$ when $x$ is rational, and let $f(x)=0$ when $x$ is irrational. Then every positive rational $r$ is a period of $f(x)$. In particular, $f(x)$ is periodic but has no shortest period.
$6.$ Quite often, the sum of two periodic functions is not periodic. For example, let $f(x)=\sin x+\cos 2\pi x$. The first term has period $2\pi$, the second has period $1$. The sum is not a period. The problem is that $1$ and $2\pi$ are incommensurable. There do not exist positive integers $a$ and $b$ such that $(a)(1)=(b)(2\pi)$.
First, let's discuss what the definition of a period is for a periodic function. A function $f$ is periodic with period $T$ means $f(t+T) = f(t)$ for all $t$.
The period of $\sin$ is $2\pi$ by definition. (You might ask why $\sin$ is defined this way, but that question may be outside the scope of this thread.) This means that $\sin(t+2\pi)=\sin(t)$ for all $t$.
Now that we know that the period of $\sin$ is, what is the period of $\sin(kt)$? Let us define a function $g$ as $g(t)=\sin(kt)$. We are asking, what is the period of $g$. That is, what value $T_g$ satisfies $g(t+T_g)=g(t)$ for all $t$.
We know $\sin(kt+2\pi)=\sin(kt)$ for all $t$. So what value of $T_g$ satisfies $k(t+T_g)=kt+2\pi$? Solving for $T_g$, we see that $T_g=\frac{2\pi}{k}$.
Best Answer
Each function's period obviously divides $2\pi$. Let's take it from there.
As Michal Zapala has noted, $f=\sqrt{2}\sin(x+\frac{\pi}{4})$ has period $2\pi$.
Since $g=\tan x\,(1+\cos x)$ cannot return to its $x=0$ value of $0$ until $\tan x=0$ at multiples of $\pi$ or $\cos x =-1$ at odd multiples of $\pi$, $\pi$ divides the period, so it's $\pi$ or $2\pi$. But $x\mapsto x+\pi$ changes the sign of $g$, so the period will have to be $2\pi$ after all.
Similarly, since $h(x)=h(0)\implies \tan x = 0\implies \pi | x$, $h$ has period $\pi$ or $2\pi$. In fact this time the period is $\pi$, since $h(x+\pi)=h(x)$.