Distribution of $X(t)=A\cos(t)+B\sin(t), t \ge 0$. $A,B$ iid $\sim N(0,1)$ Let $A, B$ be independent random variables which are both $N(0,1)$ distributed.
$X(t)=A\cos(t)+B\sin(t), t \ge 0$ is a stochastic process.

I want to determine
a) $\mathbb E[X(t)], \mathbb V[X(t)]$
b) The distribution of $X(t)$
c) The finite-dimensional distributions of $X$

a)
$E[X(t)]=E[A \cos(t)+B \sin(t)]=E[A]\cos(t)+E[B]\sin(t)=0$
$V[X(t)]=V[A \cos(t)+ B \sin(t)]=\cos^2(t)V(A)+\sin^2(t)V(B)=\cos^2(t)+\sin^2(t)=1$
b)
I know that if $A$, $B$ are independent and $A\sim$$N(\mu_1, \sigma_1^2)$, $B\sim$ $N(\mu_2, \sigma_2^2)$  distributed then $A+B$ is $N\Big(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2\Big)$ distributed.
So $X(t)$ is $N(0,2$) distributed.
Is that correct?
And for c) I have no idea how I can determine the finite-dimensional distributions of X
$\mathbb E((A \cos(s)+B \sin(s))(A \cos(t)+B \sin(t)))=\mathbb E(A^2 \cos(s)\cos(t)+AB (\cos(s)\sin(t)+\cos(t)\sin(s))+B^2 \sin(s)\sin(t))\\=\cos(s)\cos(t)\sigma^2 + \sin(s)\sin(t)\sigma^2 + (\cos(s)\sin(t)+\cos(t)\sin(s))\mathbb E(AB )=\\=\cos(s)\cos(t)+\sin(s)\sin(t)$
 A: As @kimchi lover pointed out, your answer to (b) is not correct. $X(t) \sim N(0,2)$ would mean, in particular, that $X(t)$ has variance $2$ which contradicts your calculation in (a). Your answer to part (a) is correct.
Concerning part (c): There is the following statement

A random vector $(Y_1,\ldots,Y_n)$ is Gaussian if, and only if, $$\sum_{j=1}^n \alpha_j Y_j$$ is Gaussian for any choice of $\alpha_1,\ldots,\alpha_n \in \mathbb{R}$.

Fix $n \in \mathbb{N}$ and $0 \leq t_1 < \ldots < t_n$. Since
$$\sum_{j=1}^n \alpha_j X(t_j) = A \sum_{j=1}^n \alpha_j \cos(t_j) + B \sum_{j=1}^n \alpha_j \sin(t_j)$$
and $A$ and $B$ are independent Gaussian random variables, it follows from the above lemma that the random vector $(X(t_1),\ldots,X(t_n))$ is Gaussian. The distribution of a Gaussian random vector is uniquely determined by its mean vector and covariance matrix, i.e.
$$m := \begin{pmatrix} \mathbb{E}(X(t_1)) \\ \vdots \\ \mathbb{E}(X(t_n)) \end{pmatrix} \quad \text{and} \quad C := (c_{ij}))_{i,j=1,\ldots,n} := (\mathbb{E}(X(t_i)X(t_j))_{i,j=1,\ldots,n}.$$
Your calculations for part (a) show that $m=0$ and that $c_{ii} = 1$ for all $i=1,\ldots,n$. Hence, the only missing thing is $\mathbb{E}(X(t_i) X(t_j))$ for $i \neq j$. I leave it to you to calculate it; let me know if you run into trouble. 

 Correct result: $\mathbb{E}(X(t_i) X(t_j)) = \cos(t_i-t_j)$.

